Question

a. What is the pH of 25 mM HCl?

b. What concentration of acetic acid will have a pH of 3.6? (For acetic acid, pKa=4.76)
c. What is the pH of 25 mM phosphoric acid? (For H3PO4, pKa1 = 2.12, pKa2 = 7.21, pKa3 = 12.32)

d. If 2 volumes of 30 mM NaOH are mixed with 1 volume of 25 mM phosphoric acid, what will be the pH of the mixture?

e. If 10 ?mol acetic acid is generated in a1.0 ml enzymatic reaction buffered by 50 mM Na- phosphate (pH 8.0), what will be the final pH of the reaction mixture?

Answer 1

1.   10mM = 25*10^-3 OR 25*10^-2
[H+] = 25*10^-2
pH = -log 25*10^-2
pH = 0.6020

2. Ka = 4.76E-5

. .CH3COOH <---> CH3COO- + H+
I. . .0.1. . . . . . .. . .. . .. . . .0. . . . . . 0
E. .0.1-x. . . . . . . . ....... . . .x. . . . . . .x

ka = [CH3COO-][H+] / [CH3COOH]

4.76E-5 = x^2 / 0.1 - x

x = 0.00131 M = [H+]

pH = -log[H+] = 2.88

To make the sodium acetate 0.1 M with respect to that salt, you would need 0.1 M sodium acetate. This will make the ratio of acid/conjugate base 1:1.

pH = pKa + log(A-/HA)

When the ratio is 1:1, then the pH = pKa. So the pH = 4.76.

3. pKa1 = 2.148 = [H+][H2PO4 -] / [H3PO4]
2.148 = x^2 / 0.01 - x
x^2 + 2.148x - 0.02148 = 0
x = [H2PO4-] = 0.00995M

pH = pKa + log[Na2HPO4]/[NaH2PO4]
7.55 = 7.198 + log[Na2HPO4]/[NaH2PO4]
0.352 = log[Na2HPO4]/[NaH2PO4]
2.25 = [Na2HPO4]/[NaH2PO4]
you will need 2.25 x the amount of Na2HPO4 as NaH2PO4
x / y = 2.25
x + y = 0.01

2.25y + y = 0.01
3.35y = 0.01
y = 0.003M = [NaH2PO4]
x = 0.007M = [Na2HPO4]

0.003moles NaH2PO4 x 120g/mole = 0.36g
0.007moles Na2HPO4 x 139g/mole = 0.973g

4. i assume 0.291 and 0.350 are concentration in mol/dm3

----------------------------
NaOH + HCL --> NaCL + H20

no of mol of HCL = 0.350 x (25/1000) = 8.75 x 10^-3

1 mol of HCL react with 1 mol of NaOH
8.75 x 10^-3 mol of HCL react with 8.75 x 10^-3 mol of NaOH

volume = 8.75 x 10^-3/0.291 = 30.1cm3 (3.s.f)