What is the pH when 5.0 mL of 0.010 M barium hydroxide is added to 20.0 mL of 0.020 M nitric acid

12.08

1.92

12.15

1.85

Answer is: pH is 1.92.

Balanced chemical reaction: Ba (OH) ₂ + 2HNO₃ → Ba (NO₃) ₂ + 2H₂O.

V (Ba (OH) ₂) = 5.0 mL : 1000 mL/L.

V (Ba (OH) ₂) = 0.005 L; volume of barium hydroxide.

c (Ba (OH) ₂) = 0.010 M; molarity of barium hydroxide.

n (Ba (OH) ₂) = V (Ba (OH) ₂) · c (Ba (OH) ₂).

n (Ba (OH) ₂) = 0.00005 mol; amount of barium hydroxide.

n (HNO₃) = 0.0004 mol; amoun of nitric acid.

From balanced chemical reaction: n (Ba (OH) ₂) : n (HNO₃) = 1 : 2.

Δn (HNO₃) = 0.0004 mol - 0.0001 mol.

Δn (HNO₃) = 0.0003 mol; excess of nitri acid.

Δc (HNO₃) = 0.0003 mol : 0.025 L.

Δc (HNO₃) = 0.012 M.

pH = - log (0.012 M).

pH = 1.92.