Chemistry with buffers?

A beaker with 100 ml of an acetic acid buffer with a pH of 5.00 is sitting on a benchtop. The total molarity of acid and conjugate base in this buffer is 0.100 M. A student adds 7.30 ml of a 0.360 M HCl solution to the beaker. How much will the pH change? The pKa of acetic acid is 4.760.Chemistry with buffers?
pH = pKa + log (base/acid)



5.00 = 4.76 = log (x/1)



I want the ratio of base to acid which is why I used x/1



log x = 0.24



x = 1.74



my base:salt ratio is 1.74:1



let x = moles of acid



1.74x + x = 0.01 total moles of solute



x = 0.00365 moles (this is the acid)

base = 0.00635 mol



Add strong acid, it protonates the base



(0.360 mol/L) (0.00730 L) = 0.002628 mol of strong acid added



base: 0.00635 - 0.002628 = 0.003722

acid: 0.00365 + 0.002628 = 0.006278



pH = 4.76 + log(0.003722/0.006278)



pH = 4.53



Note: I used moles in that last H-H expression rather than using the new volume of 107.3 mL to get molarities. Since there would be a 0.1073 L in the numerator and the denominator, I just skipped that step.



Good problem. Check my math carefully and talk the solution through with your study group. Best wishes.



Update:



http://www.chemteam.info/AcidBase/Hender



problem #7

Chemistry help please?

A beaker with 175 mL of an acetic acid buffer with a pH of 5.00 is sitting on a benchtop. The total molarity of acid and conjugate base in this buffer is 0.100 M . A student adds 6.90 mL of a 0.340 M HCL solution to the beaker. How much will the pH change? The pKa of acetic acid is 4.760Chemistry help please?
let x = [acetic acid]

let y = [acetate]



5.00 = 4.760 + log y/x

10^0.24 = y/x

1.74 = y/x

1.74 x = y



but we know that x + y = 0.100

y = 0.100-x

1.74 x = 0.100-x

2.74 x = 0.100

x = 0.0365 and y = 0.0635 M



moles acetic acid = 0.0365 x 0.175 L=0.00639

moles acetate = 0.0635 x 0.175 L=0.0111



moles H+ added = 6.90 x 10^-3 L x 0.340 M=0.00235

the reaction that occur is

CH3COO- + H+ = CH3COOH

moles acetate = 0.0111 - 0.00235 =0.00875

moles acetic acid = 0.00639 + 0.00235 =0.00874

total volume = 6.90 + 175 = 181.9 mL =%26gt; 0.1819 L

[acetic acid ] = 0.00874 / 0.1819 =0.0480 M

[acetate] = 0.00875/0.1819 =0.0481 M



pH = 4.760 + log 0.0481 / 0.0480=4.76

Chemistry Acid Base Buffer?

A beaker with 155 mL of an acetic acid buffer with a pH of 5.00 is sitting on a benchtop. The total molarity of acid and conjugate base in this buffer is 0.100 M. A student adds 7.60 mL of a 0.260 M HCl solution to the beaker. How much will the pH change? The pKa of acetic acid is 4.760.Chemistry Acid Base Buffer?
let x = [acetate]

let y = [acetic acid]



x + y = 0.100



5.00 = 4.760 + log x/y



5.00 - 4.760= log x/y



10^ 0.24 =1.74 = x/y



1.74 y = x



1.74 y + y = 0.100

2.74 y = 0.100

y=0.0365 M

x = 0.100 - 0.0365=0.0635 M



moles acetate = 0.0635 x 0.155 L=0.00984

moles acetic acid = 0.0365 x 0.155=0.00566



moles H+ added = 7.60 x 10^-3 L x 0.260=0.00198



CH3COO- + H+ = CH3COOH

moles acetate = 0.00984 - 0.00198=0.00786

moles acetic acid = 0.00566 + 0.00198=0.00764



total volume = 0.1626 L

[acetate]= 0.00786/ 0.1626=0.0483

[acetic acid]= 0.00764/ 0.1626=0.0470 M



pH = 4.760 + log 0.0483/ 0.0470=4.77



delta pH = 0.23

Do you think it's safe to put flagstone in a goldfish tank?

Hi,

i was wondering if it's safe to put it and build it. its a 20 gallon goldfish aquarium! a few question!

is it safe for the goldfish?

will it put to much pressure on the glass? (it's about 70 pounds)

how do i clean it fist

and will it change the PH?

is it safe for goldfish????!?!?Do you think it's safe to put flagstone in a goldfish tank?
Should be fine. Just be careful with the weight. Seventy pounds is a bit too much for a twenty gallon tank, just be sure not to put any ';points'; directly on the glass which would create a pressure crack. If you're laying it flat, no problem. I'd go with small/medium pieces, around six-eight inches, and stack them mabey halfway up the tank. A few could reach the surface. Leave the front half of the tank empty. This shouldn't be too much weight. Glass is stronger than most people think. If you have bubble-eye goldfish, be sure there are no sharp edges that could burst their bubbles! Just rinse it off in warm water to remove the dust.Do you think it's safe to put flagstone in a goldfish tank?
Why on earth would you want to do that?

Challenging Pure chemistry?

When volume of acid was added to an alkali, how does pH change during the acid -base titration?



Thank you very much for your help.Challenging Pure chemistry?
';The pH will level off';??? I don't believe that Rhiz has ever done a titration. There is no ';leveling off'; at the equivalence point. Take a look at the graph of pH vs volume of added base for the titration of an acid.

http://library.tedankara.k12.tr/chemistr

or this one for the titration of a base by an acid:

http://www.chem.ubc.ca/courseware/pH/sec



Challenging Pure chemistry?
The pH will decrease progressively...



Should this be a strong acid-strong base titration, the pH will decrease progressively then rising or descending sharply (equivalence point) along pH 7...



In a strong acid-weak base titration... the drastic descent or equivalence point will be below pH 7...
It Depends On the Buffering Capacity.
There is a non-linear response. pH changes relatively rapidly until you reach a buffering point (not really a point), a narrow pH range where the dominant compound undergoes an ion change by the addition of H+ to the base form. the pH where this occurs depends on the stability relations between the H-base and H-free base ions. You see only a slight pH change in the immediate vicinity of this point because the H+ that is added is essentially immediately combined with the base ion. after you have converted essentially all of the base to the H-base, the pH will again change rapidly.

Advice needed for sick drawf (freshwater) puffers.?

African (Malawi) cichlid tank. 75 gallons, established. PH 7.5-8.0. Chichlid salt used. No plants, only rockscape. Bought 6 puffers ~8 weeks ago. Feed mosquito larvae, brine shrimp, krill. I feed them once a day. Occasionally skip a day. Also pellet, flake, cichlid food. The puffers don't seem to eat these. No signs of illness. No problems with other fish in the tank. Found three of the puffers dead, two days ago. One is now sick. THe other two lay on the bottom but seem to have revived and are swimming around None of them have looked sick- no ich, fungus, etc. Help. I really like these guys and am not sure what to do. Is the Ph too high for them? Not enough or too muh salt? No cover? Wrong food? looked them up on the net but cannot find the answers to these questions. I have a community tank and a guppy tank I could move them to if the ph is an issue (they are 7.0)- but if so, how do you do it without shocking them by the ph change? Any suggestions would be appreciated.Advice needed for sick drawf (freshwater) puffers.?
dwarf puffers could be taking each other out -- they aren't very social and are better with heavy plants if kept in groups. from my understanding they are salt sensitive anything over a teaspoon and 1/2 in 10 gallons would be a problem.Advice needed for sick drawf (freshwater) puffers.?
Puffers should have absolutely no salt content. They are also solitary fish and would have double the stress in this big, stocked tank.
kill it. If you touch it then u will grow a fin and ure legs will be gone and then u will have a huge head and then u will eat something and it will mutate inside of u and you will become a fish!!!!!!!!!!!!!!

































jking. YA just no salt 4 him.

Decorating the bedroom need some ideas

slightest effect

Chemistry help please?

A beaker with 175 ml of an acetic acid buffer with a pH of 5.00 is sitting on a benchtop. The total molarity of acid and conjugate base in this buffer is 0.100M . A student adds 6.90 mL of a 0.340 M HCL solution to the beaker. How much will the pH change? The pKa of acetic acid is 4.760.



Basically whats delta pH?Chemistry help please?
let y = [acetate]

let x = [acetic acid]



5.00 = 4.760 + log y/x

10^0.24 = 1.74 = y/x

1.74 x = y

but we know that y + x =0.100



1.74 x = 0.100-x

x = 0.0365 M

y = 0.0635 M

moles acetic acid = 0.0365 x 0.175 L= 0.00639

moles acetate = 0.0635 x 0.175 L= 0.0111

moles H+ added = 6.90 x 10^-3 L x 0.340 M= 0.00235

CH3COO- + H+ = CH3COOH

moles acetate = 0.0111 - 0.00235 = 0.00875

moles acetic acid = 0.00639 + 0.00235 = 0.00874

pH = 4.760 + log 0.00875 / 0.00874 = 4.76



delta pH = 5.00 - 4.76=0.24

Common ion effect question?

What is the Ph change of a 0.210 M solution of citric acid pka equals 4.77 If citrate is added to a concentration of 0.150 M with no change in volume?



please does anyone have any idea on how to solve this?Common ion effect question?
well, the best way to do this would be the hendersson hasslebach equation, which is pH = pKa + log(A-/HA). in this cause, you first need to know how much of the original 0.210M citric acid dissociates in a solution. Therefore, make a chart that helps you show how much citric acid dissociates base on it's Ka1. After that, you would add 0.150M citrate. Based on the rules of equilibrium, the equation will work to balance this effect, and you need to balance the effects by changing the concentration of HA and A-, which is how you get the values for the equation.

Is a weak acid on (its own) a buffer?

When text books explain how a weak acid buffer opposes the effect of the addition of an alkali they tend to refer to how the alkali reacts with and removes H+ ions but then the weak acid deprotonates to replenish these H+ ions (Le Chatalier's principle). This explanation does not rely on or require the salt of the weak acid!



Put simply: Would a weak acid solution (without its corresponding salt) act as a buffer opposing a change in pH on addition of an alkali?Is a weak acid on (its own) a buffer?
It would oppose a change in pH on the addition of an alkali. However, a real buffer is supposed to resist all changes in pH, in response to the addition of acid also. The above is NOT an actual buffer; you need the weak acid AND it's conjugate base to be able to buffer a solution. It is an BUFFERING AGENT, A buffering agent is used to adjust or stabilize an acidic or basic solution, by CHANGING it's pH. On the other hand a buffer solution MAINTAINS the pH of the solution.Is a weak acid on (its own) a buffer?
weak acid %26amp; weak base salts are buffers....not a weak acid acts as a buffer ........ neutralization theory explains this
Have to agree with a previous post. It would oppose the change of pH when adding alkali, but there will simply not be enough of the conjugate base present (because the acid is weak!) to oppose a change in pH when acid is added.

Where does Ick come from?

I've actually put this hear to explain how the disease works. I had made a response to a previous question, and someone said they didnt know why I thought ph had anything to do with ick. Actually Ick is a parasite that all fish carry, but break out with when stressed. Usually any drastic change in ph or tempature will bring on an ick breakout.

Ich is a ciliated protozoan parasite that infests freshwater tropical fish, goldfish, koi, and other gamefish species. Ich is a relatively large protozoan, up to one mm in diameter. Ich infestations can wipe out an entire tank of fish or pond if left untreated.

The most common symptom is the appearance of white spots on the fish. The spots can be seen on fins, the body, and eyes of the fish. Infested fish may not immediately show the characteristic white spots. Ich infests the gills, feeding on cells and fluids. Gill tissue suffers extensive damage, leading to suffocation of the fish.

Ich also infests the body and fins and can lead to secondary bacterial and fungal infections. In the early stages of infection fish may be seen scratching on ornaments, rocks, or gravel. In the later stages fish are often seen hanging near power filter outlets, pumping their gills, in an attempt to get oxygen. Some fish may sit of the bottom of the aquarium or pond. Infested fish often will not eat.

Ich parasites burrow just under the skin of fish, causing the characteristic ';white spot'; or trophont stage. At maturity, the adult parasite, called a tomont, detaches from the fish and swims freely for about six hours. The trophozite eventually settles to the bottom of the aquarium. The parasite then secretes a protective membrane. The ';cyst'; now undergoes many divisions, producing 1,000 or more offspring, called theronts. When the cyst breaks open, up to 1000 theronts emerge in search of a fish host. Theronts invade their fish host by burrowing into the skin with their cilia and digestive enzymes. The tomites feed on fish cells and tissue fluids until mature, starting the cycle over again. Tomites especially devastating to delicate gill tissue. The gills are destroyed by the destructive feeding action of the parasites, causing the fish to suffocate.



Considering that each trophozite releases about 1000 infective theronts, it is easy to see how fish can quickly succumb to an Ich outbreak. Water temperature controls the speed of the Ich life cycle. At 21掳 -24掳 C (70掳 -75掳 F) it takes about three days for a complete cycle.

Ich parasites can only be killed when they are in the free-swimming theront stage. Medications do not kill the parasites attached to the fish (white spot) or when the parasites are encysted in the gravel. Disappearance of the white spots simply means that the parasites have advanced to the cyst stage. In a few hours or days, depending on water temperature, thousands of infective theronts will burst out in search of a fish host. It is precisely at this point that the medication does its job. Since not all the Ich parasites ';hatch out'; at the same time, it is necessary to treat the aquarium or pond for several days to insure control. When one fish has ick, all fish in the aquarium or pond will be infected. All fish must be treated. Ich parasites are easily transferred to other aquaria or ponds by nets, hands, boots, etc. Quarantine the infested fish. Do not add or remove fish from the infested aquarium or pond. Begin treatment immediately.Where does Ick come from?
Ich is not always present on all fish. It is not a disease. It is a parasite. It is temperature sensitive. It's life cycle is 7 days.



If you raise the temp to 86 degrees and don't see any white spots on your fish for 10 days, the parasite has died out in your tank and cannot reinfect unless you add an infected fish or a plant/decoration from an infected tank.



A holding tank for new fish and plants is a must. Two weeks is the absolute minimum to house the new fish/plant in the holding tank. Remember, 7 days is the life cycle of the ich parasite. Other diseases or parasites may present during this time also, or even in the third week. Many top authorities in the field of fish diseases and parasites reccommend a 4 week period in the holding tank before adding the newcomers to your community tank.



If you have already tried medication without results, there is another way to cure ich. The other approach is to actually destroy the organism with heat, and can be combined with the salt treatment, but not with meds.



The data that was studied (including a report by the Southern Regional Aquaculture Center) suggests that most strains of Ich cannot reproduce at temperatures above 85潞F. To use this temperature treatment approach, slowly (no more than 1 or 2 degrees per hour) raise the temperature to 86潞F, while maintaining strong continuous surface agitation to oxygenate the water.



This is extremely important because water holds less O2 at higher temperatures. (This is why meds should not be used in conjunction with high temp ?most Ich treatment products also reduce oxygen levels. Less available oxygen, combined with the respiration difficulties an infected fish is already faced with, could be fatal.)



The adjusted temperature should be maintained for approximately 10 days, or a minimum of 3 days after all signs of the parasite have disappeared (the life cycle of the parasite is 7 days).



Do not discontinue treatment when the spots go away. This is critical, because we know that the parasites are visible only as a white spot (trophont) on the body of the host, and not during the reproductive or free-swimming stage. We also know that trophonts on the gills are impossible to see.



Salt inhibits all freshwater parasites and is commonly used with many livebearers to keep them healthy. Most fish benefit from salt in their water, but some are affected detrimentally. Do your research before adding salt to any tank.Where does Ick come from?
I stand corrected.



I do however stand by the fact that different fish have different Ph requirements. But that is not the point here.



EDIT: Nevermind, this is not your own writing. http://badmanstropicalfish.com/meds/ick.
I think this is the best fish website I've seen, please join the forum
Boy lots of false information here in the question and also in the responses.



Here is a good article on Ich.



http://www.aquaria.info/index.php?name=N
first off be assured all fish do not carry ich. Ich is a free swimming parasite. The only way to totally elimate it is to have a sterilizer on your tank. Also where did you get your info? Research better. Ich can also affect saltwater fish. Perhaps you need a better web site or should I say more than one to figure out what exactly you are tryin to say. Good copy from a web page however you missed many different items. The web page where you got your info was from a novis not a pro. This is the problem with surfing the web. Don't take the info that sounds like what you want to hear, take what is correct. Ich does not attach itself to non stressed fish. NO MATTER what this web page says. Medications DO kill the free swimming form as well as a sterilizer. Where did you get this crap from? Disapperance of the white spots, right does not mean it is cured but they have not necessarily ';Moved on to another stage.'; GOD you people will believe anything. Water temps don't mean didly. Sterilize and treat, however Kick ich may be way beyond your financial means, then again, if this quack diesn't recommend it it must not be true. Quarantine a fish after the entire tank is infected? that is a smart move bowles. Each medication is different. There is no host fish, they all are treated as such. I think you need to go back to aquarium 101 hun since apparently you didn't learn anything the first time around. AND QUIT surfing web pages and taking the crack pot sites as fact. No wonder your tanks die. You should not be allowed to have fish until you learn how to actually care for them. What a waste of time and typing.
I am not sure......all I know is that it is a disease(for fish of course).

good luck!

Adding strong acid to buffer?

A beaker with 165mL of an acetic acid buffer with a pH of 5.00 is sitting on a benchtop. The total molarity of acid and conjugate base in this buffer is 0.100M . A student adds 9.00mL of a 0.250M solution to the beaker. How much will the pH change? The of acetic acid is 4.760.Adding strong acid to buffer?
0.100 M * 165 mL = 16.5 mmoles CH3COOH and CH3COONa.



I'm going to assume that you're adding 9.00 mL 0.250 M strong acid to the solution.



9.00 mL * 0.250 M = 2.25 mmoles HCl



The strong acid will react with the conjugate base form to produce more weak acid.



16.5 mmoles CH3COONa - 2.25 mmoles HCl = 14.25 mmoles CH3COONa leftover.



16.5 mmoles CH3COOH + 2.25 mmoles CH3COOH created = 18.75 mmoles CH3COOH



Use the Henderson-Hasselbalch equation to find the new pH:



pH = pKa + log[A-/HA]



You normally would convert the new mmoles to concentration, but you do not have to because this is a ratio between the two.



pH = 4.760 + log(14.25/18.75) = 4.64



Since the solution before the addition of the acid is an even mixture of acid and conjugate base, the pH = pKa initially.



So the change in pH would be:



dpH = pHf - pHi = 4.64 - 4.76 = -0.12

Adding strong acid to buffer?

A beaker with 105ml of an acetic acid buffer with a pH of 5.00 is sitting on a benchtop. The total molarity of acid and conjugate base in this buffer is 0.100M . A student adds 6.60ml of a 0.360M solution to the beaker. How much will the pH change? The pKa of acetic acid is 4.760.Adding strong acid to buffer?
5.00 - 4.76=0.24



10^0.24 =1.74 = [acetate]/ [acetic acid]



[acetate]= 1.74 x [acetic acid]



[acetate] + [acetic acid]= 0.100



[acetic acid]= 0.100- [acetate]



[acetate]= 1.74 ( 0.100 - [acetate] = 0.174 - 1.74 acetate



[acetate]= 0.0635 M

[acetic acid ]= 0.0365 M



moles acetate = 0.105 L x 0.0635 =0.00667

moles acetic acid = 0.0365 x 0.105 L=0.00383



now the student add 6.60 mL of a 0.360 M of ???

hope helps

cs files

vc

Adding strong acid to buffer?

A beaker with 105ml of an acetic acid buffer with a pH of 5.00 is sitting on a benchtop. The total molarity of acid and conjugate base in this buffer is 0.100M . A student adds 6.60ml of a 0.360M solution to the beaker. How much will the pH change? The pKa of acetic acid is 4.760.Adding strong acid to buffer?
5.00 - 4.76=0.24



10^0.24 =1.74 = [acetate]/ [acetic acid]



[acetate]= 1.74 x [acetic acid]



[acetate] + [acetic acid]= 0.100



[acetic acid]= 0.100- [acetate]



[acetate]= 1.74 ( 0.100 - [acetate] = 0.174 - 1.74 acetate



[acetate]= 0.0635 M

[acetic acid ]= 0.0365 M



moles acetate = 0.105 L x 0.0635 =0.00667

moles acetic acid = 0.0365 x 0.105 L=0.00383



now the student add 6.60 mL of a 0.360 M of ???

hope helps

Chemistry help please?

Explain what is in a buffer. Discuss the function of a buffer. How will pH change when small amounts of acids or bases are added to the buffer solution?Chemistry help please?
a pH buffer is a substance added to a reaction to keep it at a certain pH. Otherwise in some reactions they change the pH themselves thus sometimes inhibiting the reaction. For example the enzyme lipase breaks down fats into fatty acids and glycerol. But the enzyme itself can only work in an alkali condition. So when it breaks down the fats,the fatty acids would change the pH to acidic and so the enzyme would no longer work- bile is the pH buffer in our body.



So back to your question. A pH buffer keeps the pH constant at the set pH wanted by the experimenter, despite any changes in the pH caused by the reaction.Chemistry help please?
I guarantee you that this question is answered in one or two paragraphs in your textbook. Look up buffer in the index. It all has to do with the percentage of Hydrogen ions involved, hence the term p for percentage, H for Hydrogen, or pH. Think of how many positive ions it would take to neutralize all of the negative ions / or vice versa. Kinda like doing addition with positive and negative numbers in the same equation. It's in the book. Look it up.

I need help with science! please!?

if the base is added to an acidic solution, how will the pH change-increase, decrease or stay the same?I need help with science! please!?
An acidic solution is that smaller than 7pH and a basic solution is larger than 7pH. So is a base is added to an acid you get a neutralization curve and you approach neutral pH. Number gets larger.I need help with science! please!?
First, always remember A-A....Add Acids to bases, don't add bases to acids. It's a standard rule in chemistry. Anyway, the pH of the acid would increase if you add a base to it. It was low to begin with, now you're adding a base (high pH). Also, remember small p, large H....pH....not being picky, but if you run inot someone picky about these things, it might matter. Add acid, and call it pH.

Adding strong base to basic buffer solution.?

What will be the pH change when 20.0mL of 0.100 M NaOH is added to 80.0mL of a buffer solution consisting of 0.169 M NH3 and .183 M NH4CL?



I calculated that before the NaOH is added the pH is 9.29 but I don't know how to do the rest.Adding strong base to basic buffer solution.?
Okay, so what you have to do here is find the moles of NaOH and the moles of your initial acid and base, find the new molarities after the change, and use the henderson-hasselbalch equation with your new concentrations.



so for the initial pH, i used pKa of NH4Cl (9.3).



pH = 9.3 + log (0.169M/0.183M), and I found the pH to be 9.265, which is a little different.





then, I found the moles of NaOH using M = mol/L



0.1M = x mol/0.02L, and found the moles to be 0.002.



after finding the moles of NaOH, i did the same method to find the moles of NH3 and NH4Cl, which were 0.0135 and 0.01464, respectively.



when adding a strong base to a buffer, the acidic component reacts with the added base, causing OH- ions to be released and the moles of acid to decrease and base to increase, respectively.



with this knowledge, I added the number of moles of NaOH to the moles NH3

0.0135 + 0.002 = 0.0155 moles



and subtracted the number of moles of NaOH from the moles of NH4Cl

0.01464 - 0.002 = 0.01264 moles



i then found the molarities of each by dividing the moles by (0.08 L + 0.02 L to account for the addition of NaOH), and found the new concentrations.



[NH3] = 0.155M, [NH4Cl] = 0.1264M



now that we have the new concentrations of acid and base, substitute them into the H-H equation to get the new pH:



pH = 9.3 + log (0.155/0.1264) = 9.39.



9.39 - 9.265 = a pH change of +0.125.



there may be some minor errors in there, but that is how one generally figures out these problems. good luck!

Problem Help! Henderson Hasselbalch equation.

Here's the problem I am given: A beaker with 120mL of an acetic acid buffer with a pH of 5.00 is sitting on a benchtop. The total molarity of acid and conjugate base in this buffer is 0.1M. A student adds 6.60mL of a 0.300M HCl solution to the beaker. How much will the pH change? The pKa of acetic acid is 4.76.



The main thing I'm having a problem with is figuring out how to find the moles of acid and base using the henderson hasselbalch equation. I found the ratio of conjugate base to conjugate acid (1.74) but that's as far as I've gotten. If someone could show me how to do that I would really appreciate it.



Thanks for your help!Problem Help! Henderson Hasselbalch equation.
Ok I would change thew molarity as it wree to mmoles of ...in other words the 120 mL X 0.1 M = 12 mmoles total for the two forms acid and base. Since the pH is 5 then the amount of each form can be determined by th following H/H 5.0 = 4.76 + log base /12 - base since the pH is merely a function of the ratio of mmoles since the two forms are in the same volume. so 5.0 - 4.76 = log base /12-base

so 1.74 = base /12-base

20.88 = 2.74 base base = 7.62 mmoles and acid = 12 - 7.62 = 4.38 moles



check



pH = 4.76 + log 7.62/4.38 = 4.76 + log 1.739 = 4.76 + 0.24 = 5







so now the pH after adding the HCl will be driven more acidic because the ratio uis going to change 6.6mL x 0.3 M = 1.98 mmole of H+ so adding this to the buffer will convert 1.98 moles of the base form ( acetate ) to the acid form so the pH will now be



pH = 4.76 + log[ 7.62-1.98]/4.38 + 1.98 mmoles = .76 + log 5.64/6.36

= 4.76 -0.052 = 4.708 it changes by - .29 pH units

How to get rid of weeds?

I have a patch of weeds that I need to kill. I am going to put grass there so I don't want to change the pH of the soil. Is there a quick way to do this? How to get rid of weeds?
everyone is right about needing to pull them out. the easiest way to do that is to thoroughly soak the base of the weed(s) by putting a hose on it (barely turned on) for about 10 minutes, and in many cases you can just grab the weed at the base and pull it out, root and all. use a weed knife to make sure you get the entire root. don't shake the weed once you get it out, throw the entire thing away, dirt and all, to ensure you don't drop any seeds. weed killer usually puts the root into hibernation mode, and it will thrive once again with enough watering. How to get rid of weeds?
You could dig for their roots. That way, the weeds can't grow back. I suggest a hoe if it grew to an amazing height.
The safest and most efficient way is to take a shovel and dig them up by roots and all and shake the dirt back out. Some people prefer chemicals but they are bad for your soil and the bees. If it's a larger area use a rototiller and then rake out as much of the weeds as you can and throw them in a compost pile elsewhere.
The quickest way, of course, is to till the weeds under with a roto-tiller, but if you were wondering about weed killer, you could use that too without changing the pH of the soil. You have to prep the soil for the new lawn seed anyway, so...remove the big weeds and till the rest under. The new lawn should choke out the weeds once it is established.
I use an herbicide called 2-4d it won't affect grass,ponds,wildlife,or house pets but kills weeds quick.
Pour boiling water on them. They will be brown the next day, then dig them up and make sure you get all the roots.



Good Luck!
If the weeds are green %26amp; activly growing us shuld use ';Roundup'; - very effective %26amp; breaks down quickly.

rodents

acrylic nails gel nails

Please explain what is in a buffer...?

and discuss the function of a buffer.

How will pH change when small amounts of acids or bases are added to the buffer solution?Please explain what is in a buffer...?
A buffer is a solution of a weak acid and it's conjugate weak base. The pH of the buffer is such that it is within one pH unit of the pKa of the acid (aka the weak acid and weak base forms are present in near-equal amounts). This means that when you add small amounts of acid or base to the buffer, the weak acid/base will sort of ';soak up'; the extra H+ (or OH-) ions and the result will be little to no change in the pH of the solution.



So the purpose of a buffer is to make a solution where the pH stays stable even when small amounts of acid or base are added.

Adding a Strong Acid to a Buffer?

have the solution, but I still dont understand some of the steps.



A beaker with 115mL of an acetic acid buffer with a pH of 5.00 is sitting on a benchtop. The total molarity of acid and conjugate base in this buffer is 0.1M . A student adds 4.30mL of a 0.490M HCl solution to the beaker. How much will the pH change? The Pk_a of acetic acid is 4.76.





SOLUTION:



pH = pKa log acetate / acid

5.00 = 4.76 log acetate / acid



acetate / acid =10^0.24= 1.74

acetate acid = 0.1



Solve the system

acetate = 0.0365 M

acid = 0.0635 M



*HOW DID THEY FIND THAT IT'S .0365 M %26amp; .0635 M???

*CAN SOMEONE SHOW ME HOW TO THE MATH FOR IT?????



moles acetate = .115 L x 0.0365 M = 0.00420 mol

moles acid = .115 L x 0.0635 M = 0.00730 mol



Moles H added = .00430 L x 0.490 M = 0.00211mol

CH3COO- H --%26gt; CH3COOH

moles acetate = 0.00420 - 0.00211 = 0.00209 mol

moles acid = 0.00730 0.00211 = 0.00941 mol



*WHY SUBTRACT FOR ACETATE BUT THEN ADD FOR ACID??



Total volume = 115 4.30 = 119.3 mL = 0.1193 L

concentration acetate = 0.00209/0.1193 = 0.0175 M

concentration acid = 0.00941 / 0.1193 = 0.0789 M



pH = 4.76 log 0.0175 / 0.0789 = 4.11Adding a Strong Acid to a Buffer?
[CH3COO-] / [CH3COOH] = 1.74

[CH3COOH] + [CH3COO-] = 0.1



[CH3COO-] = 0.1 - [CH3COOH]



We put this value in the 1st equation :

0.1 -[CH3COOH] / [CH3COOH] = 1.74



we multuply the left and the right side by [CH3COOH]



0.1 - [CH3COOH] = 1.74 [CH3COOH]

0.1 = 2.74 [CH3COOH]

[CH3COOH] = 0.0365 M



[CH3COO-] + 0.0365 = 0.1

[CH3COO-] = 0.1 - 0.0365 = 0.0635 M



the effect of the added 0.00211 mol of H+ would be to decrease the moles of CH3COO- by 0.00211 and increase the moles of CH3COOH by 0.00211 by the reaction :

CH3COO- + H+ %26gt;%26gt; CH3COOH