Question

Using the concentration values determined in this investigation for M(OH)2 and HCl, calculate the pH for a strong acid + strong base titration in which 5.00 mL of the M(OH)2 was transferred via pipet to a beaker and HCl was added from the buret.

Concentration of base is 0.07 M

Concentration of acid is 0.0532 M

Calculate the pH:

a) before any HCl is added

b) after the addition of 4.00 mL HCl

c) after the addition of 9.00 mL HCl

d) 4.00 mL beyond the equivalence point

Answer 1

a)

M(OH)2 = M+2 + 2OH-

[OH-] = 2*[M(OH)2] = 2*(0.07) = 0.14

pH = 14 + log(OH) = 14+ log(0.14) =13.15

b)

mmol of HCl = MV = 4*0.0532 = 0.2128

mmol of M(OH)2 = MV = 5*0.07 = 0.35

mmol of OH- = 2*0.35 = 0.70

mmol of OH- left = 0.70 - 0.2128 = 0.4872

V total = 4+5= 9

[OH-] = mmol/V = 0.4872 / 9 = 0.05413 M

pH = 14 +log(OH) = 14+ log(0.05413)

pH =12.7334

c)

mmol of HCl = MV = 9*0.0532 = 0.4788

mmol of M(OH)2 = MV = 5*0.07 = 0.35

mmol of OH- = 2*0.35 = 0.70

mmol of OH- left = 0.70 - 0.4788= 0.2212

V total = 9+5= 14

[OH-] = mmol/V = 0.2212/ 14= 0.0158 M

pH = 14 +log(OH) = 14+ log(0.0158)

pH =12.19

d)

mmol of OH- = 0.70

mmol of H+ required for equvi = 0.70

V = mmol/M = 0.70/0.0532 = 13.1578 mL

V total = 13.1578 + 4 = 17.1578

mmol of H+ left = 4*0.0532 = 0.2128

[H+] = 0.2128 / 17.1578)= 0.0124

pH = -log(0.0124 = 1.906