Question

Determine the pH of a solution when 26.5 mL of 0.026 M HNO3 is mixed with 17.9 mL of:

A) 0.0070 M NaOH. pH=?

B) distilled water. pH=?

C) 0.070 M HCl. pH=?

D) 0.084 M KOH. pH=?

Answer 1

Given

26.5 ml of 0.026 M HNO3

No. of moles of HNO3 = Volume in litres * Molarity = 0.0265 L * 0.026 mol/L = 6.89 * 10^-4 moles of HNO3

HNO3 is strong acid so it will diassociate completely the diassociation reaction

HNO3 (aq) ------> H+ (aq) + NO₃^2- (aq)

So 6.89 * 10^-4 moles of HNO3 will produce 6.89 * 10^-4 moles of H+

(A) if we add 17.9 ml of 0.0070 M of NaOH

No. of moles of NaOH = 0.0179 L * 0.0070 mol/L = 1.253 * 10^-4 moles of NaOH

NaOH is strong base complete diassociation and diassociation reaction is

NaOH (aq) -----> Na+ (aq) + OH- (aq)

so 1.253 * 10^-4 moles of NaOH will give 1.253 * 10^-4 moles of OH-

H+ (aq) + OH- (aq) ----> H2O (l)

1.253 * 10^-4 moles of OH- will nuetralize 1.253 * 10^-4 moles of H+

so there will be (6.89 * 10^-4 - 1.253 * 10^-4 ) = 5.637 * 10^-4 moles of H+ will reamain

Volume of solution = 26.5 ml + 17.9 ml = 44.4 ml = 0.0444 L

[H+] = 5.637 * 10^-4 moles / 0.0444 L = 0.01269 M

pH = - log (H+) = - log ( 0.01269) = 1.89

(B) 17.9 ml of Distilled water

6.89 * 10^-4 moles of H+ will be present as there will be no H+ or OH- ions produced by water

Volume of solution = 26.5 ml + 17.9 ml = 44.4 ml = 0.0444 L

[H+] = 6.89 * 10^-4 moles / 0.0444 L = 0.015518 M

pH = - log (H+) = - log ( 0.015518) = 1.81