Question

A 100.0 mL sample of 0.10 M NH3 is titrated with 0.10 M HNO3. Determine the pH of the solution before the addition of any HNO3. The Kb of NH3 is 1.8 × 10-5.

Answer 1

Answer:

$pH=11.12$

Explanation:

Hello,

In this case, ammonia dissociation is:

$NH_3(aq)+H_2O(l)\rightleftharpoons NH_4^+(aq)+OH^-(aq)$

So the equilibrium expression:

$Kb=\frac{[NH_4^+][OH^-]}{[NH_3]}$

That in terms of the reaction extent and the initial concentration of ammonia is written as:

$1.8x10^{-5}=\frac{x*x}{0.10M-x}$

Thus, solving by using solver or quadratic equation we find:

$x=0.00133M$

Which actually equals the concentration of hydroxyl ion, therefore the pOH is computed:

$pOH=-log([OH^-])=-log(0.00133)=2.88$

And the pH from the pOH is:

$pH=14-pOH=14-2.88\\\\pH=11.12$

Best regards.