Solid $NH X_{4} Cl$ (53.5 g/mol) is dissolved in 500 mL of 0.20 M $NH X_{3}$ to give a solution of pH 9.00. Given $p K_{b} = 4.74$ , $p K_{w} = 14$ and $1 0^{0.26} = 1.82$ , find the mass of $NH X_{4} Cl$ in grams, to one decimal (no volume change).

Numerical answer type, answer in g, tolerance 0.1.

Answer: 9.7 g

pOH = 5.00 =

4.74 + lo g ([NH X_{4} X^{+}] / [NH X_{3}])

, so ratio =

1 0^{0.26}

= 1.82. NH3 = 0.100 mol, so NH4+ = 0.182 mol, mass =

0.182 \times 53.5 = 9.7