Question

Determine the pH of a solution that is 3.70% KOH by mass. Assume that the solution has density of 1.01 g/mL .

Answer 1

Let volume of solution be 1 L

volume , V = 1 L

= 1*10^3 mL

density, d = 1.01 g/mL

mass = density * volume

= 1.01 g/mL *1*10^3 mL

= 1010.0 g

This is mass of solution

mass of KOH = 3.7 % of mass of solution

= 3.7*1010.0/100

= 37.37 g

Molar mass of KOH,

MM = 1*MM(K) + 1*MM(O) + 1*MM(H)

= 1*39.1 + 1*16.0 + 1*1.008

= 56.108 g/mol

mass(KOH)= 37.37 g

number of mol of KOH,

n = mass of KOH/molar mass of KOH

=(37.37 g)/(56.108 g/mol)

= 0.666 mol

volume , V = 1 L

Molarity,

M = number of mol / volume in L

= 0.666/1

= 0.666 M

This is concentration of KOH

[OH-] = 0.666 M

use:

pOH = -log [OH-]

= -log (0.666)

= 0.1765

use:

PH = 14 - pOH

= 14 - 0.1765

= 13.8235

Answer: 13.8