Question

# Using the concentration values determined in this investigation for M(OH)2 and HCl, calculate the pH for...

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

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

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