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# Acids and Bases Watch

1. I'm struggling with some questions and I would really appreciate it if you could help me out and explain how to answer the following questions. Thank you for looking.

Q1
Determine the pH of the following;
(a) A solution containing 0.1 mol dm-3 formic acid (Ka = 1.8x10-4) and 2.0 mol dm-3 sodium formate.
I get this to be 4.05 but I don't understand the second part.

(b) 1 dm-3 solution of (a) after the addition of 5ml of 2.0 mol dm-3 HCL
(neglect any volume change)

Q2
Two buffer solutions were made from formic acid (Ka = 2.0x10-4 moldm-3) and sodium formate. In the first solution the concentrations of the acid and the salt were both 0.9 moldm-3 whereas in the second solution the concentrations of both constituents were 0.33 moldm-3.

(a) Calculate the pH of both solutions
(b) Which solution has the greater buffer capacity.

Q3

A buffer solution of pH 5 is required, the acid used has a Ka value of 1.8x10-5 and the if it's of use the ratio of acid to salt is 1.8:1.

If the solution is to be prepared by adding sodium hydroxide to a solution of the acid what percentage of the acid must be neutralised?
What volume of 0.1 mol l-1 NaOH is required per litre of 0.1 mol l-1 weak acid?
2. (Original post by Picture~Perfect)
I'm struggling with some questions and I would really appreciate it if you could help me out and explain how to answer the following questions. Thank you for looking.

Q1
Determine the pH of the following;
(a) A solution containing 0.1 mol dm-3 formic acid (Ka = 1.8x10-4) and 2.0 mol dm-3 sodium formate.
I get this to be 4.05 but I don't understand the second part.

(b) 1 dm-3 solution of (a) after the addition of 5ml of 2.0 mol dm-3 HCL
(neglect any volume change)

Q2
Two buffer solutions were made from formic acid (Ka = 2.0x10-4 moldm-3) and sodium formate. In the first solution the concentrations of the acid and the salt were both 0.9 moldm-3 whereas in the second solution the concentrations of both constituents were 0.33 moldm-3.

(a) Calculate the pH of both solutions
(b) Which solution has the greater buffer capacity.

Q3

A buffer solution of pH 5 is required, the acid used has a Ka value of 1.8x10-5 and the if it's of use the ratio of acid to salt is 1.8:1.

If the solution is to be prepared by adding sodium hydroxide to a solution of the acid what percentage of the acid must be neutralised?
What volume of 0.1 mol l-1 NaOH is required per litre of 0.1 mol l-1 weak acid?
Well for starters it seems to me you've answered 1a wrong, if we take it approximately then [H+]=Ka*Ca/Cs and pH comes as 5.05. I have checked it with more precise methods and the answer is the same.

For the second part I assume it's as simple as working out the number of moles of H+ previously (9*10-6 moles), then how much is added when you add HCl (0.01 moles), add them together (honestly let's just go with 0.01 moles) and then moles/volume for the concentration (0.01 moldm-3) and then get pH. I haven't checked this with any more precise method (sorry) but it seems this should work. pH is roughly 2.
3. (Original post by Big-Daddy)
Well for starters it seems to me you've answered 1a wrong, if we take it approximately then [H+]=Ka*Ca/Cs and pH comes as 5.05. I have checked it with more precise methods and the answer is the same.

For the second part I assume it's as simple as working out the number of moles of H+ previously (9*10-6 moles), then how much is added when you add HCl (0.01 moles), add them together (honestly let's just go with 0.01 moles) and then moles/volume for the concentration (0.01 moldm-3) and then get pH. I haven't checked this with any more precise method (sorry) but it seems this should work. pH is roughly 2.
I appreciate your input but there isn't anything obviously wrong with my answer to the first part of question one. I have the answers for question one and three. 1(a) is 4.05 and 1(b) is 3.98
I got the answer for 1(a) using the following method;
Ka=[H+][A-]/[HA]
[H+] = ka[HA]/[A-] = (1.8x10-4 x 0.1)/0.2 = 9.0x10-5
pH= -log9.0x10-5 = 4.05

I just don't understand how to apply that to part (b) if that is what your supposed to do.
4. (Original post by Picture~Perfect)
I appreciate your input but there isn't anything obviously wrong with my answer to the first part of question one. I have the answers for question one and three. 1(a) is 4.05 and 1(b) is 3.98
I got the answer for 1(a) using the following method;
Ka=[H+][A-]/[HA]
[H+] = ka[HA]/[A-] = (1.8x10-4 x 0.1)/0.2 = 9.0x10-5
pH= -log9.0x10-5 = 4.05

I just don't understand how to apply that to part (b) if that is what your supposed to do.
Why are you using 0.2 for [A-]? If we take the Cs=[A-] approximation then [A-]=2.0 according to your original question.

How did you get that answer for b)? Let me make this simple for you: there is no conjugate base for HCl present in the solution yet, so the [H+] will be at least the H+ which comes from HCl, which is at least 0.01 moldm-3 leading to a pH of at most 2.
5. I'll do the approximate calculations for 2.

(a) Calculate the pH of both solutions

How is this any different from 1a? (For which your method was fine.)

[H+]=Ka*[HA]/[A-]≈Ka*Ca/Cs

Now plug the numbers in. pH=3.70 for both solutions (of course it will be the same, since the ratio of conjugate base to acid is the same - very approximately). You aren't being asked to use a more exact method than the HH equation I presume?

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Updated: April 10, 2013
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