An 0.500*L volume of ammonia/ammonium buffer has pH=9.15. If [NH_3]=0.10*mol*L^-1, what MASS of ammonium chloride does this buffer contain?

Mar 25, 2017

We use the buffer equation, $p H = p {K}_{a} + {\log}_{10} \left\{\frac{\left[N {H}_{3}\right]}{\left[N {H}_{4}^{+}\right]}\right\}$

Explanation:

Now for $\text{ammonium ion}$ in water $p {K}_{a} = 9.24$

And thus, $9.15 = 9.24 + {\log}_{10} \left\{\frac{\left[N {H}_{3}\right]}{\left[N {H}_{4}^{+}\right]}\right\}$

Clearly, ${\log}_{10} \left\{\frac{\left[N {H}_{3}\right]}{\left[N {H}_{4}^{+}\right]}\right\} = - 0.09$

And so, $\frac{\left[N {H}_{3}\right]}{\left[N {H}_{4}^{+}\right]} = {10}^{- 0.09} = 0.813$

And given that $\left[N {H}_{3}\right] = 0.10 \cdot m o l \cdot {L}^{-} 1$, $\left[N {H}_{4}^{+}\right] = 0.123 \cdot m o l \cdot {L}^{-} 1$.

Because the volume of the solution was specified to be $0.500 \cdot L$, there were $0.500 \cdot L \times 0.123 \cdot m o l \cdot {L}^{-} 1 = 0.0615 \cdot m o l$, i,e, $0.0615 \cdot m o l \times 53.49 \cdot g \cdot m o {l}^{-} 1 = 3.29 \cdot g$.

For the derivation of the $\text{buffer equation}$, see here for details.