Question

What is the pH of this solution?

Answer 1

Answer:

pH = 11.216.

Explanation:

Hello there!

In this case, according to the ionization of ammonia in aqueous solution:

$NH_3+H_2O\rightleftharpoons NH_4^++OH^-$

We can set up its equilibrium expression in terms of x as the reaction extent equal to the concentration of each product at equilibrium:

$Kb=\frac{[NH_4^+][OH^-]}{[NH_3]} \\\\1.80x10^{-5}=\frac{x*x}{0.150-x}$

However, since Kb<<<1 we can neglect the x on bottom and easily compute it via:

$1.80x10^{-5}=\frac{x*x}{0.150}\\\\x=\sqrt{1.80x10^{-5}*0.150}=1.643x10^{-3}M$

Which is also:

$[OH^-]=1.643x10^{-3}M$

Thereafter we can compute the pOH first:

$pOH=-log(1.643x10^{-3}M)\\\\pOH=2.784$

Finally, the pH turns out:

$pH=14-2.784\\\\pH=11.216$

Regards!