Question

Calculate the pH of a solution prepared by mixing 25.0 mL of 0.100M nitric acid with 25.0 mL of 1.00M barium hydroxide. (Indicate the limiting reactant and reactant in excess below.

Answer 1

2HNO₃ (aq.) + Ba(OH)₂ (aq.) -------------> Ba(NO₃)₂ (aq.) + 2 H₂O (l)

Given that,

Moles of HNO3 = 0.100 * 25.0 / 1000 = 0.00250 mol

Moles of Ba(OH)2 = 1.00 * 25.0 / 1000 = 0.0250 mol

From the above balanced equation,

2 mol HNO3 requires 1 mol Ba(OH)2

Then, 0.00250 mol HNO3 requires 1 * 0.00250 / 2 = 0.00125 mol of Ba(OH)2

SO, HNO3 is limiting reagent. Because, 0.00125 < 0.025

Remaing moles of Ba(OH)2 = 0.025 - 0.00125 = 0.02375 mol

Therefore, new concentration of Ba(OH)2 = 0.02375 / 50.0 = 0.000475 M

Ba(OH)2 (aq.) -----------> Ba2+ (aq.) + 2 OH- (aq.)

[OH-] = 2 * 0.000475 = 0.000950 M

we know that,

pH + pOH = 14

pH = 14 - pOH

pH = 14 + Log[OH-]

pH = 14 + Log(0.000950)

pH = 10.98