Question

Determine the pH of a solution that is 3.55% KOH by mass. Assume that the solution has density of 1.01 g/mL.

Answer 1

Given the percent of KOH is 3.55 %

Thatmeans 3.55 g of KOH present in 100 g of solution.

Given density of solution , d = 1.01 g/mL

So volume of the solution is V = mass / density

                                            = 100 g / 1.01 (g/mL)

                                            = 99.01 mL

                                            = 99.01 / 1000 L       Since 1L = 1000 mL

                                            = 0.09901L

Molar mass of KOH = 39 + 16 + 1 = 56 g/mol

So numb er of moles of KOH , n = mass/molar mass

                                               = 3.55 g / 56 (g/mol)

                                              = 0.063 mol

Molarity of KOH = number of moles / volume of the solution in L

                      = 0.063 mol / 0.09901 L                      

                      = 0.636 M

KOH ----> K⁺ + OH^-

1 mole of KOH produces 1 mole of OH^-

So [OH^-] = [KOH] = 0.636 M

pOH = - log[OH^-]

       = - log 0.636

       = 0.19

We know that pH + pOH = 14

So pH = 14 - pOH

         = 14 - 0.19

        = 13.81

Therefore the pH of the solution is 13.81