By mixing $ 500 \; mL \ NH_3 \; 0.2 \ M $ to $ 500 \; mL \; NH_4Cl (aq) $ we obtain a solution of $ pH = 9.0$ Let us calculate the mass of ammonium chloride used: Taking $ x $ $ = $ $ n_{NH_4Cl} $ $ = $ $ n_{NH_4^+} $ (Each mole of ammonium chloride produced by dissolving one mole of ammonium) ...
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Equation ?
$pH$ $=$ $pK_a+log\frac{n_{NH_3}}{n_{NH_4^+}}$ $9.0$ $=$ $9.20+log\frac{0.5\cdot 0.2}{x}$
Resolution?
$log\frac{0.1}{x}$ $=$ $-0.20$ $\frac{0.1}{x}$ $=$ $10^{-0.20}$ $x$ $=$ $\frac{0.1}{10^{-0.20}}$ $=$ $0.16\; mol$
Now calculate the mass of ammonium chloride!
$m_{NH_4Cl}$ $=$ $n_{NH_4Cl}M_{NH_4Cl}$ $=$ $0.16\cdot 53.5$ $=$ $8.56\;g$
