Someone good with practicals would get this!?

i'm doing a bio project, testing the effect of pH on immoblized and non-immoblized enzymes, using invertase enzyme and sucrose solution. BUT the only way to test for glucose is through a diabetes strip thing, however when changing pH i will get different colours( ie pH1= red...etc) and when i pour this pH 1 solution into the enzyme/sucrose solution it will change colour. After that i would put the strip inside the test tube, but i will not see the colour change of the strip with this different colour (made by the pH) and i would not be able to tell how much glucose there is in there!!! HELPPP, HOW DO I FIX THIS PROBLEM!!



thanks for your help, sorry about the explanation!Someone good with practicals would get this!?
add the acidic ph, take the new reading, then neutralize it with water.

What is the PH of a solution that contains 0.15M hc2h3o4 and 0.25 M c2h3o2? use Ka= 1.8 X 10-5?

what is the PH of a solution that contains 0.15M hc2h3o4 and 0.25 M c2h3o2? use Ka= 1.8 X 10-5.....................i know how to do this part.... Ph is 3.3



i need to do second part: by how much will the Ph change if 0.025 mol of HCL is added to 1.00 L of the buffer in problem (i have stated up_What is the PH of a solution that contains 0.15M hc2h3o4 and 0.25 M c2h3o2? use Ka= 1.8 X 10-5?
CH3COO- + H+ %26gt;%26gt; CH3COOH

moles acetate = 0.25 - 0.025 =0.225

[CH3COO-]= 0.225 / 1.00 L= 0.225

Moles acetic acid = 0.15 + 0.025=0.175

[CH3COOH]= 0.175 / 1.00L = 0.175

pKa = 4.7



pH = pKa + log [CH3COO-]/ [CH3COOH]

pH = 4.7 + log 0.225 / 0.175=4.8



about the first question :

pH = 4.7 + log 0.25/ 0.15= 4.9

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What is the pH of final solution?

A beaker with 185 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 7.20 mL of a 0.300 M HCl solution to the beaker. How much will the pH change? The pKa of acetic acid is 4.760.



Express your answer numerically to two decimal places. Use a minus (-) sign if the pH has decreased.What is the pH of final solution?
First use the Henderson-Hasselbalch equation to calculate the concentrations of acetic acid and acetate in the initial buffer:



pH = pKa + log [acetate]/[acetic acid]

5.00 = 4.76 + log [acetate]/[acetic acid]

log [acetate]/[acetic acid] = 0.24

[acetate]/[acetic acid] = 1.74

[acetate] = 1.74 [acetic acid] **



Now, you know that [acetic acid] + [acetate] = 0.100 M. It's often easier to work in just moles rather than in actual molar concentrations, so multiplying the total concentration by the volume of the buffer solution, you have:

acetic acid + acetate = 0.0185 moles



Substituting the starred equation into the second one gives:

acetic acid + 1.74 acetic acid = 0.0185

2.74 acetic acid = 0.0185

moles acetic acid = 6.75 X 10^-3 moles and then moles acetate = 0.0185 - 6.75 X 10^-3 = 0.01175 moles.



Now, if you add 7.20 mL of 0.300 M HCl, you have added 2.16 X 10^-3 moles of HCl.



When HCl is added to a buffer, the moles of acetate will decrease by exactly the amount of HCl added, and the moles of acetic acid will increase by that same amount. So, after the HCl addition,



moles acetic acid = 6.75 X 10^-3 + 2.16 X 10^-3 = 8.91 X 10^-3 moles

moles of acetate = 0.01175 - 2.16 X 10^-3 = 9.59 X 10^-3 moles.



Now, use the H-H equation to calculate the final pH:



pH = 4.760 + log 9.59X 10^-3 / 8.91 X 10^-3 = 4.79.



So, the pH dropped by 0.21 pH units.

Chemistry Help with Adding a Strong Acid to a Buffer-trying to find delta pH.?

Okay so here is the problem:



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 9.00 mL of a 0.410 M HCl solution to the beaker. How much will the pH change? The pKa of acetic acid is 4.760. What is the delta pH? (change in pH).





What I initially thought--

H C2H3O2 --%26gt; H+ %26amp; (C2H3O2)-1

I .005549____________.009951

C +.00419 ____________-.00419

E .009739____________.005761

(btw I got .00419 by adding .009 and .410 then dividing by 1000)



then I used this formula:

1.74x10^-5= x(.005761/.009739)

x=[H+]= 2.94148x10^-5

-log(2.94148x10^-5)= 4.5313

therefore the pH=4.53

since they were asking for the change of pH I subtracted

5.00-4.53 = .4686

and apparently that is INCORRECT....



ssooo.....I dont know what I did wrong

PLEASE help me, and give step my step instructions as well

Because I'm totally lost now...Chemistry Help with Adding a Strong Acid to a Buffer-trying to find delta pH.?
Let the initial concentrations be a and b(of acid and conjugate base res.)



5 = 4.76 + log(b/a)



So b/a =1.738.



And b + a = 0.1



Solving the two, b = 0.0365 and a = 0.0635



milliMoles of acid = 0.0635*155 = 9.8425



milliMoles of conjugate base = 5.6575



On addition of .410*9 = 3.69 milliMoles HCl, the acid = 9.8425 + 3.69= 13.5325 and base = 5.6575 - 3.69 = 1.9675.



So the pH = 4.76 + log(1.9675/13.5325) = 4.76 - 0.8375



So change in pH is 5 - 4.76 + 0.8375 = 1.0775(approx.)



There might have been some calculation mistake please check the calculations.

Is ocean ';acidification'; really caused by CO2?

The ocean contains 50 times as much CO2 as the air. The CO2 has increase about 1/3 for rounding purposes in the air. One third of the increased CO2 could be attributed to humans. Of that, 1/3 has went into the sea. 1/3 or 1/3 is 1/9 so approximately an amount equivalent to 1/9 of the atmospheric CO2 went into the ocean. The ocean contains 50 times as much CO2. That means the the increase in the ocean is roughly 1/9 of approximately 1/50 or we have changed the concentration of the CO2 in the ocean by about 1/450 The roughness of the approximation and rounding down of the 1/3 increase makes it seem reasonable to round that up to to 0.25% or 4 parts per thousand. Can you get significant acidification from a weak acid in buffer conditions with that small an increase or is it just alarmism as usual trying to lie and frighten people by suggesting we are acidifying the ocean. Carbonate ion changes with temperature and biological activity changes pH to as does volcanoes and other factors. We can easily figure how much a change of that magnitude would have on the pH. It isn't significant, that much is certain.



Before you respond with something like, ';It is more than the environment can handle'; or some similar nonsense, tale a deep breath and consider how stupid equating a 4 parts per thousand increase in CO2 to doom is. There used to be over 17 times as much CO2 in the air. I am a chemist so please don't distort science and claim that the concentration of the H+ ion which is what acid measures increased 35%. The change was from 8.2 to 8.1. That is very roughly a 10% change since pH as most high school science students know is logarithmic. Acid is below 7 and I think I have demonstrated with simple undeniable logic, except perhaps to an alarmist, that it is ignorant to attribute the acid change to CO2.Is ocean ';acidification'; really caused by CO2?
OK, your additional details redeemed at least the first part of your post. I was about to lay it on you for that. The rest of your playing with fractions really is irrelevant. Why not just start pulling out the gelatinous math and say the oceans are only absorbing .0001% of the total atmosphere?



In a basic solution like the ocean carbonic acid deprotonates almost completely into HCO3- (alpha 1 = .975, or 97.5% deprotonation). So while it is technically a weak acid, in the basic ocean it is pretty strong.



And, Mr. Chemist, you need to review your algebra. 10^.1 = 1.26, or a 26% increase.



Finally, no one is saying that the oceans are going to become like acid that will burn the flesh off of fish. The main problem, rather, is that increasing the acidity causes carbonate concentration [CO3-2] to fall below the levels of saturation, reducing the rates of calcification of many primary producers. It is this inhibition of phytoplankton's ability to produce aragonite that is the main cause for concern, not the pH itself--though this may have some nasty effects as well.

http://www.ipsl.jussieu.fr/~jomce/acidif

http://www.ucar.edu/communications/FinalIs ocean ';acidification'; really caused by CO2?
Oh. Yeah we could have kept going, but alright. Thanks for BA

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Not at all. I think that toxic chemicals being released into the air (not carbon dioxide) and which then falls down it what's causing the acidic oceans (assuming they're truly more acid). It's the same thing that's causing acidic snow, rain, and water supplies that no one talks about!



http://www.youtube.com/watch?v=MKjnLeFPB



I don't get the CO2 claim--they say even carbonated water is very weak in acid (and the ocean are definitely not that extreme). And another thing is that when the oceans and absorb CO2 it actually results in carbonated mineral deposits like calcium carbonate--which is an antacid! So that means the opposite happens!



';The Earth's oceans dissolve a major amount of carbon dioxide. The resulting carbonate anions bind to cations present in sea water such as Ca2+ and Mg2+ to form deposits of limestone and dolomite. Most carbon dioxide in the atmosphere eventually undergoes this fate: if all the carbonate rocks in the earth's crust were to be converted back in to carbon dioxide, the resulting carbon dioxide would weigh 40 times as much as the rest of the atmosphere.';



http://knowledgerush.com/kr/encyclopedia



Springs full of CO2 is actually healing and good for combatting acidity problems!



';Europeans also know that carbonated Vichy Water along with its high and balanced mineral content relieves those with extreme circulatory problems. Vichy waters have unique characteristics when consumed including the relief of stomach acidity, it's effect is like a natural Alka Seltzer. It is used for stomach ulcers, rheumatism and arthritis.';



http://www.vichysprings.com/vichy_baths/



That's why I reject the CO2 link to global warming--everything they claim ends up being the opposite of the truth!
';Ocean acidification'; is a misnomer. The ocean is not acidic and won't be any time soon.



http://wattsupwiththat.com/2009/01/31/oc
Ah yes. Here again we have the crank who knows more than the combined efforts of the worlds' foremost scientists.



With a back of the envelope calculation I can prove the science is bogus!



And hey, I got no alternate explanation for the observed phenomena, and I don't need to!



Cause' I can whip out my favorite bogeyman and claim the whole thing is a communist plot!



Do you have any idea how ridiculous this makes you look?



You may want to pull your punches because you're hitting yourself in the head.



http://www.ocean-acidification.net/

Can you check if I am right with this pH question!?

This is pretty easy (I think) but I'm trying to get perfect on this assignment, so could you check my reasoning?



The range of blood pH that is considered normal in humans is 7.35 to 7.45. If a person's blood pH changes from 7.35 to 7.45, what is the change in hydronium ion concentration ([H30+])?



a) 1.3 * 10^-8 M

b) 9.2 * 10^-9 M

c) -9.2 * 10^-9 M

d) -1.3 * 10^-8 M

e) -2.3 * 10^-8 M



I get (c) and I found this by actually calculating the hydronium ion concentration at each end of the acceptable pH range and finding the difference. This was done by taking ten to the power of the negative value of the pH. Since an increase in pH results in a decrease in hydronium ion concentration, the answer should be negative. Can you check if I am right!?



ALSO, I tried it this way and it didn't work, and I'm curious why: how come it doesnt work out if you just subtract the pH values and then take the ten to the power of the negative difference, to find the H ion concentration? (do you get understand that? haha)Can you check if I am right with this pH question!?
[H+] at pH 7.35 = 10^-7.35 = 4.46*10^-8

[H+] at pH 7.45 = 10^-7.45 = 3.54*10^-8



Difference = 9.2*10^-9 answer is b)



The difference is always quoted as positive. Think about this:

what is the difference between 2 an 4? answer =2

What is the difference between 4 and 2 answer : = 2

Nowhere do you say negative,



Also Answer: Remember that pH is a logarithmic scale: You will know from your mathematics classes that if you subtract logarithms, you are not finding the difference, but you are dividing. That is why you get an incorrect answer.



If you divide 4.46*10^-8 by 3.54*10^-8 you get 1.259



If you subtract 7.45-7.35 you get 0.1. = 10^0.1 = 1.259



This latter method is not the way to do this type of problem because you are not finding the difference.Can you check if I am right with this pH question!?
Yes. pH of 7.35 is [H+] = 4.47*10^-8;

pH of 7.45 is [H+] = 3.55*10^-8.

Difference is c.



You cannot use your second technique because subtracting logs does not do the same thing as subtracting the exponentials; it is the result of dividing.



10^2 - 10^1 =(100-10) = 90.



Take logs and use your technique gives you 2-1 = 1 which raised to the tenth power is 10 which is not the same thing at all but which is 100/10.
yes, that's correct.

i thought i had to divide -9.2 * 10^-9 by 3 considering there's 3 H in H3O+, but i assumed that there's only 1 H+ and the rest is water there?

..

and ur 2nd question, of course it doesn't work.

is 10^(-7.45 - (-7.35)) same with 10^(-7.45) - 10^(-7.35)?

lol.

xD
/////////////// definitions



pH = -LOG([H(+)])



';pH equals the negative LOG of the Hydrogen ion concentration, expressed in terms of moles per liter';



///////////// calculations



7.35 = -LOG([H(+)])

-7.35 = LOG([H(+)])

10^(-7.35) = [H(+)]

[H(+)] = 4.46E-8



7.45 = -LOG([H(+)])

-7.45 = LOG([H(+)])

10^(-7.45) = [H(+)]

[H(+)] = 3.54E-8



DELTA([H(+)]) = 3.54E-8 - 4.46E-8 = -9.2E-9



The change is actually a decease, so the answer is negative, so ';c'; is correct. Higher pH values mean a decrease in the hydrogen ion concentration. However, there is no such thing as an actual negative concentration. What has been calculated is a change in pH. This is sort of like working with electrical engineering where voltage is the difference between two points in a circuit, not a single value in itself.



///////////////// how it work



You can't subtract LOG values and have them behave like ordinary numbers. On a LOG scale, subtraction is equivalent to division and addition of LOGS is multiplying the numbers they represent.



For example, LOG(100) - LOG(10) = 2-1 = 1. 10^(1) = 10 and 100/10 also = 10.



LOG(100) + LOG(10) = 2 + 1 = 3. 100*10 = 1000. LOG(1000) = 3.



Not too hard once you get the haing of this sort of thing.



FYI: The natural log function is just what happens when 1/x is integrated. Nice to know what natural logs actually represent. Too bad they never actually explained this in college.

Neon Tetras and water quality tolerance?

I've had an assortment of tetras in a 29g tank for about a year now in RO water. After a recent addition of 6 more, bringing my total up to about 35, I noticed they started to look stressed. I did a quick check of the water and found the pH was at 5.0. Was a bit shocked by this. After 5 days of adding in a small amount of dechlorinated tap water I got the pH to 6.5 and everyone started looking better. I keep up with my weekly water changes. Hence my concern. The pH going for a good 6.8 to 5.0 (or less) in under a week is not something I want to happen.



After this little episode, I am considering adapting them to tap water, which is buffered against pH change... however, the parameters aren't exactly ideal from what everything I've researched tells me neons prefer. The reason I'm considering this now is they are going to be moving into a new 55g tank in a few weeks, so if I am going to adapt them to tap water, now is the time to do it.



Current water conditions in their tank is:

GH 4

KH 1

pH 6.5



Water conditions out of the tap are:

GH 11

KH 9

pH 8



As you can see, it's not likely tap water will change pH values on me. My question to those with experience with neon tetras is: How will they do in the tap water in my area? Please, only answer if you have experience with neons. Thanks!Neon Tetras and water quality tolerance?
Thats the problem with using pure RO water in a tank. It's TOO pure.



The hardness is very low, and that means it takes very little acidic material to crash the pH.



Adding a little treated tap water as you have done is the best option. Mix back a little tap water (maybe 20%?) with your RO water to get the pH you want and it will be more stable as there will now be some hardness to act as a buffer.



IanNeon Tetras and water quality tolerance?
You have two options;

Either use a buffer solution designed to make RO water optimum for amazon species or what |i would do is mix RO water with dechlorinated tap water.

Try two parts RO to one part tap and premix it before adding.

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