Question

What is the pH of the solution that results from adding 42.8 mL of 0.150 M HCℓ to 23.9 mL of 0.650 M NH₃?

Answer 1

$HCl_{(aq)}+NH_{3(aq)} \to NH_{4(aq)}^++Cl_{(aq)}^-$

The ratio = 1:1 (meaning equal number of moles)

$c=\frac{n}{V}$ $\to n=c×V$

$n_{HCl}=\frac{0.150×42.8}{1000}=0.00642moles$

$n_{NH_3}=\frac{0.650×23.9}{1000}=0.0155moles$

Since you have twice as many moles of ammonia than you have of hydrochloric acid, you can conclude that the acid will act as a limiting reagent, i.e. it will be completely consumed by the reaction.

Now, since you have a 1:1 mole ratio between both reactants and the ammonium ions, $NH^+_4$ , the conjugate acid of ammonia, it follows that the reaction will produce as many moles as you have moles of reactants taking part in the reaction.

So, if 0.00642 moles of hydrochloric acid will react with 0.00642 moles of ammonia, the reaction will produce

$NH_4^+moles = 0.0155-0.00642=0.00908$

Total volume $=42.8+23.9=66.7ml=0.0667l$

$Molarity =\frac{0.00642}{0.0667}=0.096M$

At room temperature Kb of ammonia is listed as

$Kb=1.8×10^{-5}$

$pOH=-log(1.8×10^{-5})=4.74$

$pH=14-4.74=9.26$