Edexcel IAL Chemistry 14.20 Buffer preparation calculations

calculation of $\left[\mathrm{H}^{+}\right]$

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- rearrangement of $K_{\mathrm{a}}$ equation

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- substitution of values

and

calculation of $\left[\mathrm{RCOO}^{-}\right]=$moles required

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OR Use of Henderson-Hasselbalch

- quotation of equation

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- substitution of values

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- calculation of $\left[\mathrm{RCOO}^{-}\right]=$moles required

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Example of calculation
$$\left[\mathrm{H}^{+}\right]=10^{-3.71}=1.9498 \times 10^{-4}\left(\mathrm{~mol} \mathrm{dm}^{-3}\right)$$$$\left[\mathrm{RCOO}^{-}\right]=\frac{K_{\mathrm{a}} \mathrm{x}[\mathrm{RCOOH}]}{\left[\mathrm{H}^{+}\right]}$$$$\begin{aligned} {\left[\mathrm{RCOO}^{-}\right] } =\frac{1.55 \times 1.38 \times 10^{-4}}{1.9498 \times 10^{-4}} =1.0970 / 1.10 / 1.1\left(\mathrm{~mol} / \mathrm{mol} \mathrm{dm}^{-3}\right) \end{aligned}$$
& $=1.0970 / 1.10 / 1.1\left(\mathrm{~mol} / \mathrm{mol} \mathrm{dm}^{-}\right)$$\mathrm{pH}=\mathrm{p} K_{\mathrm{a}}+\underset{[\mathrm{RCOOH}]}{\log \left[\mathrm{RCOO}^{-}\right]}$ & $\begin{array}{l}\text { Ignore round brackets } \text { or no brackets }\end{array}$

&
$$\mathrm{pH}=\mathrm{p} K_{\mathrm{a}}+\underset{[\mathrm{RCOOH}]}{\log \left[\mathrm{RCOO}^{-}\right]} \quad \begin{aligned} \text { Ignore round brackets } \text { or no brackets } \end{aligned}$$$3.71=3.86+\log \frac{\left[\mathrm{RCOO}^{-}\right]}{1.55}$$\left[\mathrm{RCOO}^{-}\right]=1.0970 / 1.10 / 1.1\left(\mathrm{~mol} / \mathrm{mol} \mathrm{dm}^{-3}\right)$
Henderson-Hasselbalch equation with acid and salt
concentration reversed gives 2.19 mol scores (2)
Correct answer with some working scores