# Given 250*mL of 0.200*mol*L^-1 aqueous ammonia, how much ammonium chloride would be added to achieve a pH-=8.90?

Apr 25, 2017

Doubtless you mean $\text{ammonium chloride}$, $N {H}_{4} C l$. I calculate a mass of $5.85 \cdot g$..................

#### Explanation:

We use the buffer equation.......which is derived here, https://socratic.org/questions/how-do-buffers-maintain-ph#270129.

But here $\left[{A}^{-}\right] = \left[N {H}_{3}\right]$, and $\left[H A\right] = \left[N {H}_{4} C l\right]$.

For such a buffer $p H = p {K}_{a} + {\log}_{10} \left\{\frac{\left[N {H}_{3} \left(a q\right)\right]}{\left[N {H}_{4} C l \left(a q\right)\right]}\right\}$

And I also ASSUME, that the required $p H = 8.90$

Now $p {K}_{a} = 9.24$ for ammonium ion.

And thus, substituting these values into the equation..........

${\log}_{10} \left\{\frac{\left[N {H}_{3} \left(a q\right)\right]}{\left[N {H}_{4} C l \left(a q\right)\right]}\right\} = - 0.34$

i.e. $\frac{\left[N {H}_{3} \left(a q\right)\right]}{\left[N {H}_{4} C l \left(a q\right)\right]} = {10}^{- 0.34}$,

i.e. $\text{moles of ammonium chloride"xx0.457="moles of ammonia}$, BECAUSE the volume was the same in each case.

$\text{moles of ammonia} = 0.250 \cdot L \times 0.200 \cdot m o l \cdot {L}^{-} 1 = 5.00 \times {10}^{-} 2 \cdot m o l$

$\text{ammonium chloride} = \frac{5.00 \times {10}^{-} 2 \cdot m o l}{0.457} = 0.109 \cdot m o l$

And thus we have to add............

$0.109 \cdot m o l \times 53.49 \cdot g \cdot m o {l}^{-} 1 = 5.85 \cdot g$

To the initial $250.0 \cdot m L$ volume of ammonia.

Just as a recheck, let us substitute these values back into the buffer equation:

$p H = 9.24 + {\log}_{10} \left\{\frac{5.00 \times {10}^{-} 2 \cdot m o l \times \frac{1}{0.2500} \cdot L}{\frac{5.85 \cdot g}{53.49 \cdot g \cdot m o {l}^{-} 1} \times \frac{1}{0.2500} \cdot L}\right\}$

$= 8.90$ as required..........