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.

here is table 2-2:

country

percent of eligible voters that voted

rank

united states

44.9%

138

canada

60.1%

107

costa rica

84.6%

14

hungary

66.9%

82

japan

57.0%

118

egypt

27.7%

159

south africa

85.5%

10

ans is no, the data set cannot be plotted as a single circle graph