2. Consider a solution of H2CO3 where Ka1 = 4.5 x 10-7 and Ka2 = 4.7 x 10-11. Fi
ID: 1015470 • Letter: 2
Question
2. Consider a solution of H2CO3 where Ka1 = 4.5 x 10-7 and Ka2 = 4.7 x 10-11. Find the pH and concentrations of H2CO3, HCO3 - , CO3 2- in (a) 0.200 M H2CO3 (b) 0.200 M HCO3 - (c) 0.200 M CO3 2-
3. You are required to make 1 L of a 0.055 M phosphate buffer of pH 7.32 and you have at your disposal phosphoric acid (MM = 98 g/mol, pKa1=2.12, pKa2=7.21, and pKa3=12.67) and 0.0500 M NaOH. a. Calculate the amount of phosphoric acid and sodium hydroxide you would need to make this buffer. b. Outline the steps you would take to prepare this buffer c. Sketch the speciation diagram for phosphoric acid. You do not need to calculate anything, roughly sketch the curve and fill in any values you can from the data provided.
Explanation / Answer
2.
(a) For 0.2 M H2CO3
H2CO3 <==> H+ + HCO3- .... Ka1 = [H+][HCO3-]/[H2CO3]
let x amount has dissociated
4.5 x 10^-7 = x^2/0.2
x = [H+] = 3 x 10^-4 M
pH = -log[H+] = 3.523
[H2CO3] = 0.2 - 3 x 10^-4 = 0.1997 M
HCO3- <==> H+ + CO3^2- ...... Ka2 = [H+][CO3^2-]/[HCO3-]
let x amount has dissociated
4.7 x 10^-11 = x^2/0.1997
[HCO3-] = 0.1997 M
x = [CO3^2-] = 3.06 x 10^-6 M
(b) For 0.2 M HCO3-
HCO3- + H2O <==> H2CO3 + OH- .... Kb1 = [OH-][H2CO3]/[HCO3-]
let x amount has reacted
1 x 10^-14/4.5 x 10^-7 = x^2/0.2
x = [H2CO3] = [OH-] = 6.66 x 10^-5 M
pOH = -log[OH-] = 4.17
pH = 14 - pOH = 9.83
[HCO3-] = 0.2 - 6.66 x 10^-5 = 0.1999 M
HCO3- <==> H+ + CO3^2- ...... Ka2 = [H+][CO3^2-]/[HCO3-]
let x amount has dissociated
4.7 x 10^-11 = x^2/0.2
x = [CO3^2-] = 3.066 x 10^-6 M
(c) For 0.2 M CO3^2-
CO3^2- + H2O <==> OH- + HCO3- .... Kb2 = [OH-][HCO3-]/[CO3^2-]
let x amount has reacted
1 x 10^-14/4.7 x 10^-11 = x^2/0.2
x = [OH-] = [HCO3-] = 6.5 x 10^-3 M
[CO3^2-] = 0.1935 M
pOH = -log[OH-] = 2.18
pH = 14 - pOH = 11.81
HCO3- + H2O <==> OH- + H2CO3 ...... Kb1 = [OH-][H2CO3]/[HCO3-]
let x amount has reacted
1 x 10^-14/4.5 x 10^-7 = x^2/6.5 x 10^-3
[H2CO3] = 1.20 x 10^-5 M
[HCO3-] = 6.49 x 10^-3 M
3. Using Hendersen-Haselbalck equation,
pH = pKa + log(base/acid)
let x amount if base is added to form buffer
7.32 = 7.21 + log(x/0.055 M x 1 L - x)
0.071 - 1.29x = x
x = 0.071/2.29 = 0.031 mol
So moles of [HPO4^2-] = 0.031 mol
(a) moles of [H2PO4-] = 0.024 mol
380 ml of 0.055 M NaH2PO4
Volume of NaOH to be added = 0.031 mol/0.05 M = 0.62 L = 620 ml
(b) Take 380 ml of 0.055 M NaH2PO4 and mix with 620 ml of 0.05 M NaOH to prepared the buffer
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