Question

What is the pH of a solution prepared by mixing 43 mL of 0.050 M Ba(OH)2 and 38 mL of 0.35 M KOH?

Answer 1

Hello, since M = n/V you can use this equation, please note that volume must be in LITERS:

n₁ = M x V

n₁ = 0.050 x 0.043 L

n₁= 2.15 x 10^-3 mol of Ba(OH)₂ resulting in 2 times that quantity for OH^- -------> 4.3 x 10^-3 mol of OH^-

for the second compound:

n₂= M x V

n₂ = 0.35 x 0.038 L

n₂ = 0.0133 mol of KOH (and, of course OH^-)

For the answer:

M_s = n_OH/ V_s

M_s = (4.3 x 10^-3 + 0.0133) / (0.043 L + 0.038 L)

M_s = 0.217 ------> this result represents the molarity of OH^- which you can use to calculate pOH and subsequently pH

pOH = - log (OH^-)

pOH = - log (0.217)

pOH = 0.663

Finally for pH:

pH = 14 - pOH

pH = 14 - 0.663

pH = 13.33