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How would you calculate the pH of the buffer if 1.0mL of 5.0M NaOH is added to 20.0mL of this buffer?

Can someone please explain to me how to do B and C step by step so I could understand it clearly? :)

Say, for example, that you had prepared a buffer in which you mixed 10.30 g of sodium hydrogen phosphate, Na2HPO4, with 100.0 mL of 1.0 M sodium dihydrogen phosphate, NaH2PO4.

A. Calculate the pH of this buffer.
B. Calculate the resulting pH if 1.0 mL of 5.0 M NaOH is added to 20.0 mL of this buffer.
C. Calculate the mL of HCl that would have to be added to this buffer in order to lower the pH by exactly 1.0 unit.

Im not sure if I did A correctly, but...
Ka = 3.6x10^-13
pKa = -log(3.6x10,^-13)
= 12

10.30g Na2HPO4 x (1 mol Na2HPO4/ 141.96g Na2HPO4)
= 0.7256 mol Na2HPO4 (1.0 mol/1 L) x .1L
= .1 mol NaH2PO4
pH = 12 + log(.1 mol NaH2PO4/.07256 mol Na2HPO4)
= 12.14

For B. How exactly would I write the chemical equation and start the ICF table? And definitely need help on C. :(

Thank you soo much in advance.

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Airin

  1. No, A uses pK2 not pK3. And I think it is pK2 + log (BASE/ACID) but H2PO4 is the acid and HPO4 is the base. I calculate approx 8.1 but you should be more careful.
    We'll take your values as ok.
    0.7256 mol base
    0.1 mol acid
    So in 20 mL you will have 0.145 mol base and 0.02 mol acid.

    For B. add 1mL of 5M NaOH = 0.005 mols
    .........H2PO4^- + OH^- ==> HPO4^- + H2O
    I.........0.02.......0....0.145
    add..............0.005............
    C......-0.005..-0.005.......+0.005
    E.......0.015.......0........0.150

    Substitute the E line into the HH equation and solve for pH.

    C. I assume that is lowering of 1 pH unit for the solution in part A and not part B. pH lowered by 1 unit makes it about 7.1 from 8.1
    ..........HPO4^- + H^+ ==> H2PO4^-
    I.......0.7256.....0........0.1
    add HCl.............x.............
    C..........-x......-x.......+x..
    E........0.7256-x....0.......0.1+x

    Substitute the E line into the HH equation and solve for x = mols HCl added. You want to calculate mL of ? M (you didn't give that along with not giving pK2.
    7.1 = pK2 + log (0.7256-x)/(0.1+x)
    and solve for x = mols HCl
    Then M = mols HCl/L HCl
    You know mols and M (not given in the post), solve for L and convert to mL.
    Post your work if you get stuck. Check my work carefully, there are some contradictions in the problem and I may not have interpreted the problem correctly.

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    DrBob222

  2. Oh right, the parent acid was H3PO4. In part A, when calculating the pH, what exactly is pKa2? I am still unsure how to find the Ka/Kb on the tables of Ka's and Kb's.

    In part B, you mentioned that in 20 mL I will have 0.145 mol base and 0.02 mol acid.

    The calculation for 0.2 mol acid is:
    20 mL x (1 L / 1,000 mL x 1.0 mol / 1 L)
    = 0.02 mol acid
    right?

    I am a bit confused how you got 0.145 mol base.
    Is it:
    .7256 mol base x .1 L = 7.256 M
    7.256 M x.02 L
    = 0.14512 mol base

    Also, for 0.005 mol of NaOH, in which you got, would it not be 0.0005 mol NaOH?
    Is it not 0.5 M x .001 L
    = 0.00005 mol OH^-?

    Thank you. :)

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    Airin

  4. For part C, I just wanted to make sure if I did the calculations correctly when taking the anti-log.

    7.07 = 7.21 + log (0.7256-x)/(0.1+x)
    -0.14 = log (0.7256-x)/(0.1+x)
    10^-0.14 = (0.7256-x)/(0.1+x)
    (10^-0.14)(0.1+x) = (0.7256-x)
    x = 0.9018 mol HCl

    M of HCl. I'm assuming I am using the pH from part A? Although...this may be the wrong way to go about it...
    [H+] = Ka * [HPO4-] / [H2PO4]
    [H+] = 10^(-pH)
    [H+] = 10^-8.07
    [H+] = 8.51^-9 M

    [HPO4-] = [H+] * [H2PO4] / Ka
    = 8.51^-9 M * 1.0 M / 6.14x10^-8
    = 0.1386 M HCl

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    Airin

  5. I found out that Molarity of HCl was 0.5 M.

    So then, 0.5 M HCl = 0.9018 mol HCl/ L HCl
    L HCl = (0.9018 mol/0.5 M)
    HCl = 1.8036 L x 1000 mL/1 L)
    = 1803.6 mL
    = 1800 mL HCl

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    Airin

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