Question

What is the pH of the resulting solution when 22.43 mL 0.1472 M Ba(OH)2 is added to 48.39 mL 0.1623 M HCl? Enter your answer to three significant figures.

Answer 1

moles of Ba(OH)₂ = 22.43 × 0.1472 / 1000

= 3.30 × 10^-3 mol.

moles of HCl = 48.39 × 0.1623 / 1000 .

= 7.85 × 10^-3 mol.

1 mol .of Ba(OH)₂ react with 2 mol of HCl .

So, Reacting mol of HCl = 6.60 × 10^-3 .

Rest mol of HCl = 7.85 × 10^-3 - 6.60 × 10^-3

Rest mol of HCl = 1.25 × 10^-3 .

Concentration of HCl =

1.25 × 10^-3/ 77.82 × 10^-3.

= 1.25 / 77.82 = 0.01606 M.

P^H = - log concentration of H⁺.

P^H = - log 0.01606

= - (-1.78 ) .

= + 1.78 .