Question

what is the pH of a solution obtaining by adding 32.1g of NaOH and 56.3g of KOH to enough water to make 1.75l of a solution

Answer 1

Moles NaOH = 31.1 g/ 40 g/mol= 0.778

moles KOH = 56.3 / 56.107 g/mol= 1.00

total OH- = 0.778 + 1.00 = 1.778

[OH-]= 1.778 / 1.75 = 1.02 M

pOH = - 0.0086