Question

Calculate the pH and pOH for the following solutions:

a) 0.0450 M NaOH

b) 0.160 M Ca(OH)2

c) a 1:1 mixture of 0.0125 M HCl and 0.0125 M Ca(OH)2

d) a 2:3 mixture of 0.0125 M HNO3 and 0.0125 M KOH

Answer 1

pOH = -log(0.045)

pOH = 1.3468 then pH = 14 - 1.3468

pH = 12.653

pOH of Ca(OH)2 = - log (0.16)

pOH = 0.7958

pH = 14 - pOH = 14 - 0.7958

pH = 13.204

2 HCl + Ca(OH)2 -------> CaCl2 + 2 H2O

assume volume ithe solution is 1 L

number of moles of HCl = 0.0125 moles

number of moles of Ca(OH)2 = 0.0125 moles

2 moles of HCl require 1 mole of Ca(OH)2

0.0125 moles of HCl requires (1 / 2) * 0.0125 moles of Ca(OH)2

                                                       = 0.00625

number of moles of Ca(OH)2 remaining in th mixture = (0.0125 - 0.00625)

                                                                                                 = 0.0625

pOH = -log (0.00625)

pOH = 2.2041

pH = 14 - 2.2041

pH = 11.7959

HNO3 + KOH -------> KNO3 + H2O

assume given ratio as volumes of the solutions then,

number of moles of HNO3 = 0.0125 *2 = 0.025 mole

number of moles of KOH = 0.0125 * 3 = 0.0375 moles

1 mole of HNO3 requires 1 mole of KOH then ,

remaining moles of KOH = 0.0375 - 0.0125 = 0.025

pOH = -log(0.025) = 1.602

pH = 14 - pOH = 14 - 1.602 = 12.398