Hey there! Sign in to join this conversationNew here? Join for free
    • Study Helper
    • Thread Starter
    Offline

    19
    ReputationRep:
    Study Helper
    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?
    Offline

    2
    ReputationRep:
    (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.
    • Study Helper
    • Thread Starter
    Offline

    19
    ReputationRep:
    Study Helper
    (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.
    Offline

    2
    ReputationRep:
    (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.
    Offline

    2
    ReputationRep:
    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?
 
 
 
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • Poll
    Brussels sprouts
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

    Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

    Quick reply
    Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.