Question

5. A solution is prepared by mixing 0.050 mole of NH, and 0.20 mole of HCI in sufficient water to give a final volume of 500 mL. Calculate the ph

Answer 1

Answer:

pH = 0.52

Explanation:

Let's consider the following balanced equation.

NH₃ + HCl ⇄ NH₄Cl

Since the reaction is 1 mole of NH₃ to 1 mole of HCl, if we have 0.050 moles of NH₃ and 0.20 moles of HCl, the former will be the limiting reactant and the latter will be in excess. The pH will be defined by the excess of HCl which is:

moles HCl in excess = moles HCl initial - moles HCl that react = 0.20 mol - 0.050 mol = 0.15 mol

The concentration of HCl is:

$[HCl]=\frac{0.15mol}{0.500L} =0.30M$

HCl is a strong acid, so the concentration of H⁺ will also be 0.30 M. Then,

pH = -log [H⁺] = -log (0.30) = 0.52