Question

11. Calculate the pH of a solution prepared by mixing

a) 48.0 mL of 0.212 M NaOH and 66.0 mL of 0.187 M HCl

b) 69.0 mL of 0.345 M KOH and 125 mL of 0.400 M HC3H5O2

c) 350 mL of 0.130 M NH3 and 400 mL of 0.100 M HCl

Answer 1

11)

a)

millimoles of NaOH = 48 x 0.212 = 10.176

millimoles of HCl = 66 x 0.187 = 12.342

here millimoles of acid > millimoles of base

[H+] = 12.342 - 10.176 / 48 + 66 = 0.019 M

pH = -log [H+] = -log (0.019)

pH = 1.72

b) 69.0 mL of 0.345 M KOH and 125 mL of 0.400 M HC3H5O2

millimoles of KOH = 69 x 0.345 = 23.805

millimoles of CH3COOH = 125 x 0.4 = 50

CH3COOH   +   KOH   --------------------> CH3COOK + H2O

   50                  23.805                                     0               0

   26.195                0                                        23.805

pH = pKa + log [salt / acid]

     = 4.74 + lof [23.805 / 29.195]

pH = 4.70

c) 350 mL of 0.130 M NH3 and 400 mL of 0.100 M HCl

millimoles of NH3 = 350 x 0.130 = 45.5

millimoles of HCl = 400 x 0.1 = 40

NH3   +   HCl   -----------------> NH4Cl

45.5        40                                  0

5.5            0                               40

pOH = pKb + log [salt / base]

       = 4.74 + log [40 / 5.5]

pOH = 5.60

pH = 8.40