How do I calculate the change in pH, following (i) the addition of 500 ml of 1M Ba(OH)2 to 500 ml of 2O, and (ii) the addition of 500 ml of 1 M acetic acid (Ka = 1.8 x10-5) to 500 ml of H2O.How Do I calculate the pH of water after adding 500 ml of 1 M acetic acid (Ka = 1.8 x10-5) to 500 ml of H2O?
Ka= [H+]^2/[CH3COOH] so [H+]=sqrt([0.5]xKa) then pH=-log[H+]
Monday, November 22, 2010
Sperm changes the ph level in my vagina. Which creates a foul odor and discharge. How can I prevent that?
Is there something that I can take daily to prevent this? Or is there something I can do to control the ph levels of my vagina? I know that absence of sperm would be the answer, but what if I'm trying to concieve?Sperm changes the ph level in my vagina. Which creates a foul odor and discharge. How can I prevent that?
well i would say use condoms but since you're trying to conceive that wouldn't work. i know some women douche but that really isn't helpful because it messes with the natural pH too. so i guess just take a shower after you have sex and wash down there very well, being careful not to get any soap in your vagina because that could cause irritation. hope that helps!Sperm changes the ph level in my vagina. Which creates a foul odor and discharge. How can I prevent that?
your partner may need his sperm tested ,,
well i would say use condoms but since you're trying to conceive that wouldn't work. i know some women douche but that really isn't helpful because it messes with the natural pH too. so i guess just take a shower after you have sex and wash down there very well, being careful not to get any soap in your vagina because that could cause irritation. hope that helps!Sperm changes the ph level in my vagina. Which creates a foul odor and discharge. How can I prevent that?
your partner may need his sperm tested ,,
Change in pH + temperature + equilibruim problem?
need help with the following problem:
For neutral water the pH is 7.0 at 298K and 7.4 at 280K.
a) the value for Kw at 298 is 10^-14. what is the value for kw at 280K?
b) what effect does increase in temperature have on the equilibrium of dissociation of water?
c) From the data above, deduce whether the reation is exothermic or endothermic.
i think i can do b) and c) if i knew how to do part a). please help?Change in pH + temperature + equilibruim problem?
part (a)-
pH = 7.4 so [H+] = 3.98x10^-8 (pH = -log[H+])
[H+]=[OH-]
Kw = [H+][OH-]
Kw = (3.98x10^-8)^2 = 1.58x10^-15
Kw = 1.58x10^-15
For neutral water the pH is 7.0 at 298K and 7.4 at 280K.
a) the value for Kw at 298 is 10^-14. what is the value for kw at 280K?
b) what effect does increase in temperature have on the equilibrium of dissociation of water?
c) From the data above, deduce whether the reation is exothermic or endothermic.
i think i can do b) and c) if i knew how to do part a). please help?Change in pH + temperature + equilibruim problem?
part (a)-
pH = 7.4 so [H+] = 3.98x10^-8 (pH = -log[H+])
[H+]=[OH-]
Kw = [H+][OH-]
Kw = (3.98x10^-8)^2 = 1.58x10^-15
Kw = 1.58x10^-15
How to use equations to show how a buffer resists changes in pH.?
How do I show equations on how a lactate buffer resists changes in pH? The components of the lactate buffer are lactic acid, HC3H5O3, and lactate ion, C3H5O3-, often provided as sodium salt.
I know it's probably very simple, but I can't find an example, so I'm not sure exactly what it's asking. Thanks so much!How to use equations to show how a buffer resists changes in pH.?
The acid is actually in an aqueous solution which means initially, it is presented in water.
Then the acid and water reacts to create H3O+ and C3H5O3-. When you add in sodium lactate ion, NaC3H5O3-, you are basically adding in C3H5O3- to the product side into the C3H5O3- from the reaction of acid+water.
This means that the equilibrium will shift back to the left since you add in more products. When it shifts back, it sort of cancels out the reactions going right caused by the reactants hence, not losing any H+ ions, hence resists change in pH.
Now to write it out:
HC3H5O3(aq) + H2O(l) %26lt;==%26gt; H+(aq) + C3H5O3-(aq)
note: H+ = H3O+ and the Na salt is a spectator ion so we don't write it out.
And the Equilibrium equation would look like:
Kc = [C3H5O3-][H+]/[HC3H5O3]
= (C3H5O3- + x)(x)/(HC3H5O3-x)
Which can be simplified as:
Kc = (C3H5O3-)(x)/(HC3H5O3)
note: we assume x is very small so we take it out and the reason it is small is because in buffer we add x to to product side to cancel out the x in the reactant side.
Look, I know it's confusing, just take time to practice these problems and ask questions when you are stuck and you will succeed. Good Luck.How to use equations to show how a buffer resists changes in pH.?
You can easy to calculate pH of the buffer by using the
For the buffer solution you can use the Henderson-Hasselbach equation:
pH = pKa + log ([A-] / [HA] )
according that equation above pH of the buffer is determined by the ratio of ([A-] / [HA] ) so we must concern with this ratio to show how a buffer resists changes in pH.
The concentration of A- can changes if we add acid to the buffer solution as this reaction:
C3H5O3Na + H+ -%26gt; HC3H5O3 + Na+
This action makes the ratio of ([A-] / [HA] decrease, because the concentration of HC3H5O3 increases.
And the concentration of HA can changes too if we add base to the solution as this reaction
HC3H5O3 + OH- -%26gt; C3H5O3- + H2O
This action makes the ratio of [A-] / [HA] increase because the concentration of C3H5O3- increases.
As thsese examples
[C3H5O3-] / [HC3H5O3] initial = (0.50)/(0.50) = 1
When you add acid to the buffer = (0.49)/(0.51)= 0.96
When you add base to the buffer = (0.51)/(0.49)=1.04
The change in the ratio [C3H5O3-] / [HC3H5O3] is very small thus the [H+] and the pH remains constant.
hopely my answer can help you...
All you can really put is:
Ka = ([C3H5O3-][H+] )/ [HC3H5O3]
pH = -log [H+]
[H+] = [C3H5O3-] / (Ka x [HC3H5O3])
pH = - log ([C3H5O3-] / (Ka x [HC3H5O3]))
I think that should do.
HC3H5O3 ---%26gt; C3H5O3- + H+
If you look up the Ka value of lactic acid you will notice it is small. This means that the vast majority of its molecules in solutiona re undissociated. Hardly any molecules ionise to form lactate ions and hydrogen ions. So [HC3H5O3] is high.
The Sodium lactate is a soluble salt. it ionises completely.
C3H5O3-Na+---%26gt; C3H5O3- + Na+
So [C3H5O3-] is high.
If we add acid to a buffer solution, the solution counteracts this by removing them.
HC3H5O3 === C3H5O3- + H+
The position of the equilibrium shifts to the left. Sodium lactate provides the solution with lactate ions which 'mop up' the H+ ions added. Forming lactic acid and resisting changing in pH.
If we add alkali, OH- ions, these react with any H+ ions in the solution (which come from the lactic acid dissociating).
H+ + OH- --%26gt; H20
This shift the equilibrium to the right to ''replace' H+ ions.
We have plent of undisociated HC3H5O3 molecules available because its a weak acid. These are ready to split up and replace the lost H+ ions.
(In this --%26gt; are normal arrows, and === are equilibrium arrows)
Hope this helps.
This is a typical exam question, and all they require is you to show lactic acid reacting with OH-, and lactate reating with H+.
Two balanced equations...
I know it's probably very simple, but I can't find an example, so I'm not sure exactly what it's asking. Thanks so much!How to use equations to show how a buffer resists changes in pH.?
The acid is actually in an aqueous solution which means initially, it is presented in water.
Then the acid and water reacts to create H3O+ and C3H5O3-. When you add in sodium lactate ion, NaC3H5O3-, you are basically adding in C3H5O3- to the product side into the C3H5O3- from the reaction of acid+water.
This means that the equilibrium will shift back to the left since you add in more products. When it shifts back, it sort of cancels out the reactions going right caused by the reactants hence, not losing any H+ ions, hence resists change in pH.
Now to write it out:
HC3H5O3(aq) + H2O(l) %26lt;==%26gt; H+(aq) + C3H5O3-(aq)
note: H+ = H3O+ and the Na salt is a spectator ion so we don't write it out.
And the Equilibrium equation would look like:
Kc = [C3H5O3-][H+]/[HC3H5O3]
= (C3H5O3- + x)(x)/(HC3H5O3-x)
Which can be simplified as:
Kc = (C3H5O3-)(x)/(HC3H5O3)
note: we assume x is very small so we take it out and the reason it is small is because in buffer we add x to to product side to cancel out the x in the reactant side.
Look, I know it's confusing, just take time to practice these problems and ask questions when you are stuck and you will succeed. Good Luck.How to use equations to show how a buffer resists changes in pH.?
You can easy to calculate pH of the buffer by using the
For the buffer solution you can use the Henderson-Hasselbach equation:
pH = pKa + log ([A-] / [HA] )
according that equation above pH of the buffer is determined by the ratio of ([A-] / [HA] ) so we must concern with this ratio to show how a buffer resists changes in pH.
The concentration of A- can changes if we add acid to the buffer solution as this reaction:
C3H5O3Na + H+ -%26gt; HC3H5O3 + Na+
This action makes the ratio of ([A-] / [HA] decrease, because the concentration of HC3H5O3 increases.
And the concentration of HA can changes too if we add base to the solution as this reaction
HC3H5O3 + OH- -%26gt; C3H5O3- + H2O
This action makes the ratio of [A-] / [HA] increase because the concentration of C3H5O3- increases.
As thsese examples
[C3H5O3-] / [HC3H5O3] initial = (0.50)/(0.50) = 1
When you add acid to the buffer = (0.49)/(0.51)= 0.96
When you add base to the buffer = (0.51)/(0.49)=1.04
The change in the ratio [C3H5O3-] / [HC3H5O3] is very small thus the [H+] and the pH remains constant.
hopely my answer can help you...
All you can really put is:
Ka = ([C3H5O3-][H+] )/ [HC3H5O3]
pH = -log [H+]
[H+] = [C3H5O3-] / (Ka x [HC3H5O3])
pH = - log ([C3H5O3-] / (Ka x [HC3H5O3]))
I think that should do.
HC3H5O3 ---%26gt; C3H5O3- + H+
If you look up the Ka value of lactic acid you will notice it is small. This means that the vast majority of its molecules in solutiona re undissociated. Hardly any molecules ionise to form lactate ions and hydrogen ions. So [HC3H5O3] is high.
The Sodium lactate is a soluble salt. it ionises completely.
C3H5O3-Na+---%26gt; C3H5O3- + Na+
So [C3H5O3-] is high.
If we add acid to a buffer solution, the solution counteracts this by removing them.
HC3H5O3 === C3H5O3- + H+
The position of the equilibrium shifts to the left. Sodium lactate provides the solution with lactate ions which 'mop up' the H+ ions added. Forming lactic acid and resisting changing in pH.
If we add alkali, OH- ions, these react with any H+ ions in the solution (which come from the lactic acid dissociating).
H+ + OH- --%26gt; H20
This shift the equilibrium to the right to ''replace' H+ ions.
We have plent of undisociated HC3H5O3 molecules available because its a weak acid. These are ready to split up and replace the lost H+ ions.
(In this --%26gt; are normal arrows, and === are equilibrium arrows)
Hope this helps.
This is a typical exam question, and all they require is you to show lactic acid reacting with OH-, and lactate reating with H+.
Two balanced equations...
How does water hardness change?
I was thinking about water hardness for my fish and I had no clue if the hardness could change by itself. I don't have any buffers to change the pH, gH or kH. Can the kH and gH change without any buffers? I have fish in there too would that change anything either? How about pH? If so what's a good way to make it lower or higher?How does water hardness change?
Simplest?....don't worry about it.
Unless you are keeping ';wild caught'; fish that are sensitive to a set range you don't need to concern yourself as all common species (even African cichlids and Discus from opposite extremes on hardness) are tank-raised and farmed in water chemistry far removed from their country cousins and extremely adaptable to what comes out of a wide array of faucets.STABILITY of water chemistry is more important than jacking around for an ';ideal perfection';.... and stability comes from routine water changes (which you should be doing anyway,right?).
';I was thinking about water hardness for my fish ....'; .I take that to mean they are already in there,so a little late to think about it,eh? If they are doing fine,and you are doing a proper water change routine,then no reason to worry with it.How does water hardness change?
yes they will change over time due to many factors, to raise kh which is related to gh i believe, u can make a solution of baking soda(or powder whichever comes in the yellow box) and water and add that. if u want to lower ur ph use regular baking soda if u dont want to lower ph; bake the baking powder first for about an hour at 300f. ur tank will change all values over time depending on what u do to it, the best thing to do is test ur water and buy a buffer to raise the ph if need be
Please please please do not use baking soda in your fish tank. This is what is used for humane fish euthanasia = dead fish!
I can't help with gH or kH, but pH will go down as Alkalinity is raised, and vice versa. There are chemicals available, and cichlid sands, if you're looking for a higher pH.
As for hardness, I generally make the water softer by adding salt. Most city water is already plenty hard.
pouring concrete mix in the water should help to harden itfinction as one pc Dog drinking too much
Simplest?....don't worry about it.
Unless you are keeping ';wild caught'; fish that are sensitive to a set range you don't need to concern yourself as all common species (even African cichlids and Discus from opposite extremes on hardness) are tank-raised and farmed in water chemistry far removed from their country cousins and extremely adaptable to what comes out of a wide array of faucets.STABILITY of water chemistry is more important than jacking around for an ';ideal perfection';.... and stability comes from routine water changes (which you should be doing anyway,right?).
';I was thinking about water hardness for my fish ....'; .I take that to mean they are already in there,so a little late to think about it,eh? If they are doing fine,and you are doing a proper water change routine,then no reason to worry with it.How does water hardness change?
yes they will change over time due to many factors, to raise kh which is related to gh i believe, u can make a solution of baking soda(or powder whichever comes in the yellow box) and water and add that. if u want to lower ur ph use regular baking soda if u dont want to lower ph; bake the baking powder first for about an hour at 300f. ur tank will change all values over time depending on what u do to it, the best thing to do is test ur water and buy a buffer to raise the ph if need be
Please please please do not use baking soda in your fish tank. This is what is used for humane fish euthanasia = dead fish!
I can't help with gH or kH, but pH will go down as Alkalinity is raised, and vice versa. There are chemicals available, and cichlid sands, if you're looking for a higher pH.
As for hardness, I generally make the water softer by adding salt. Most city water is already plenty hard.
pouring concrete mix in the water should help to harden it
Help with ap chem lab 19: pH properties of Buffer solutions?
Calculate the pH change when 1 mL of 0.2 M HCl is added to 50 mL of deionized water. How does this pH value change compare to those obtained when 1 mL of 0.2 M HCl is added to the buffers?
At what point did each of the buffers lose their effectiveness? ExplainHelp with ap chem lab 19: pH properties of Buffer solutions?
1 mL of 0.2 M HCl is added to 50 mL of deionized water:
0.001 litres @ 0.2mol/litre = 0.0002 moles HCl
0.0002 moles HCl / 0.051 litrers = 0.0039 Molar
0.0039 Molar H+ =
pH of 2.4
that's a drastic change in pH from deionized water of about 7.0
==========================
when the same amount of HCl is added to a buffer the pH hardly drops at all, be cause the conjugate base of the buffer mix removes the added H+ ... combining with it in an equilibrium shift making a weak conjugate acid of that base
I don't have your lab exp, but this is what typpically happens
At what point did each of the buffers lose their effectiveness? ExplainHelp with ap chem lab 19: pH properties of Buffer solutions?
1 mL of 0.2 M HCl is added to 50 mL of deionized water:
0.001 litres @ 0.2mol/litre = 0.0002 moles HCl
0.0002 moles HCl / 0.051 litrers = 0.0039 Molar
0.0039 Molar H+ =
pH of 2.4
that's a drastic change in pH from deionized water of about 7.0
==========================
when the same amount of HCl is added to a buffer the pH hardly drops at all, be cause the conjugate base of the buffer mix removes the added H+ ... combining with it in an equilibrium shift making a weak conjugate acid of that base
I don't have your lab exp, but this is what typpically happens
I'd like to change my ph No on yahoo?
i will always sign in to my mobile device before i signout yahoo messenger but now i want to signin to my new number and im unable to change the number.as i click on messenger to sign out i can only sign out of yahoo but i cant sign in to my new number plz help me.and how can i sign in to my new numberI'd like to change my ph No on yahoo?
You can edit/change/delete your mobile devices from here:
http://sites.mobile.yahoo.com/wireless/m
You can edit/change/delete your mobile devices from here:
http://sites.mobile.yahoo.com/wireless/m
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