How much ammonium chloride do we need to add to a 20*mL volume of NH_3(aq) at 0.50*mol*L^-1 concentration, to maintain a pH=9.0?

Mar 8, 2017

We need (i) approx. $270 \cdot m g$, and (ii) we need to use the buffer equation: $p H = p {K}_{a} + {\log}_{10} \left(\frac{\left[N {H}_{4}^{+}\right]}{\left[N {H}_{3}\right]}\right)$.

I do not think the given answer is kosher.

Explanation:

Now we know that $p {K}_{a} + p {K}_{b} = 14$ for an aqueous solution under standard conditions, ${K}_{b} = 2 \times {10}^{-} 5$, and thus $p {K}_{b} = 4.70$, and $p {K}_{a} = 9.30$

And we substitute the $p {K}_{a}$ value back into the original equation, with the desired $p H$.

And thus,

$9 = 9.30 + {\log}_{10} \left(\frac{\left[N {H}_{4}^{+}\right]}{\left[N {H}_{3}\right]}\right)$

And thus $- 0.30 = {\log}_{10} \left(\frac{\left[N {H}_{4}^{+}\right]}{\left[N {H}_{3}\right]}\right)$

${10}^{- 0.3} = \frac{\left[N {H}_{4}^{+}\right]}{\left[N {H}_{3}\right]}$

And we know that $\left[N {H}_{3}\right] = 0.50 \cdot m o l \cdot {L}^{-} 1$.

So $\left[N {H}_{4}^{+}\right] = {10}^{- 0.3} \times 0.50 \cdot m o l \cdot {L}^{-} 1 = 0.25 \cdot m o l \cdot {L}^{-} 1$

Now $\text{concentration}$ $=$ $\left(\text{Moles of solute")/("Volume of solution}\right)$, but the volume of solution was SPECIFIED to be $20 \times {10}^{-} 3 L$.

So $\text{moles of solute} = 20 \times {10}^{-} 3 \cancel{L} \times 0.25 \cdot m o l \cdot \cancel{{L}^{-} 1}$

$= 5.01 \times {10}^{-} 3 \cdot m o l$ of $N {H}_{4} C l$; and this represents a mass of

$5.01 \times {10}^{-} 3 \cdot m o l \times 53.49 \cdot g \cdot m o {l}^{-} 1 = 0.268 \cdot g$.

So a final check,

$\left[N {H}_{4}^{+}\right] = \frac{\frac{0.268 \cdot g}{53.49 \cdot g \cdot m o {l}^{-} 1}}{20 \times {10}^{-} 3 \cdot L} = 0.251 \cdot m o l \cdot {L}^{-} 1$.

$\left[N {H}_{3}\right] = 0.50 \cdot m o l \cdot {L}^{-} 1$.

Resubstituting into the buffer equation:

$p H = p {K}_{a} + {\log}_{10} \left(\frac{\left(0.251 \cdot m o l \cdot {L}^{-} 1\right)}{\left(0.50 \cdot m o l \cdot {L}^{-} 1\right)}\right)$

$= 9.30 + \left(- 0.3\right)$

$= 9.0$ AS REQUIRED....................