Question

How many grams of ${\mathbf{Na}}_{\mathbf{2}}{\mathbf{CO}}_{\mathbf{3}}$(FM 105.99) should be mixed with $\mathbf{5}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{g}$of $\mathbf{NaHC}\mathbf{}{\mathbf{O}}_{\mathbf{3}}$(FM 84.01) to produce $\mathbf{100}\mathbf{}\mathbf{mL}$of buffer with $\mathbf{pH}\mathbf{10}\mathbf{.}\mathbf{00}$?

Short answer

The required amount to produce$100\mathrm{mL}\mathrm{of}\mathrm{buffer}\mathrm{with}\mathrm{pH}10.00\mathrm{is}\mathrm{m}=2.96\mathrm{g}$

Step-by-step solution

Definition of Henderson Hasselbalch  equation

• The Henderson Hasselbalch equation is a rough formula that demonstrates the relationship between a solution's pH or pOH, the pKa or pKb, and the ratio of dissociated chemical species concentrations.

Determine the value of pH at 10.00

It can see from the challenge that we have the carbonic acid buffer, for which two Henderson-Hasselbalch equations may be written:

$$\begin{gathered} \mathrm{pH}={\mathrm{pK}}_1+\log \frac{\left[{{\mathrm{HCO}}_3}^{-}\right]}{\left[{\mathrm{H}}_2{\mathrm{CO}}_3\right]} \\ \mathrm{pH}={\mathrm{pK}}_2+\log \frac{\left[{{\mathrm{CO}}_3}^{2-}\right]}{\left[{\mathrm{H}}_2{{\mathrm{CO}}^{-}}_3\right]} \\ \mathrm{It}\mathrm{can}\mathrm{rewrite}\mathrm{the}\mathrm{equation}\mathrm{as}\mathrm{follows}: \\ \log \frac{\left[{\mathrm{Na}}_2{\mathrm{CO}}_3\right]}{\left[{\mathrm{NaHCO}}_3\right]}=\mathrm{pH}-{\mathrm{pK}}_2 \\ \frac{\left[{\mathrm{Na}}_2{\mathrm{CO}}_3\right]}{\left[{\mathrm{NaHCO}}_3\right]}={10}^{\mathrm{pH}-{\mathrm{pK}}_2} \\ {\mathrm{Na}}_2{\mathrm{CO}}_3={10}^{\mathrm{pH}-{\mathrm{pK}}_2}.\left({\mathrm{NaHCO}}_3\right) \\ m\mathit{\left(}{\mathrm{N}}_{\mathrm{a}}{\mathrm{CO}}_3\right)={10}^{\mathrm{pH}-{\mathrm{pK}}_2}.\frac{\mathrm{m}\left({\mathrm{NaHCO}}_3\right)}{\mathrm{M}\left({\mathrm{NaHCO}}_3\right)}.M\left({\mathrm{Na}}_2{\mathrm{CO}}_3\right) \\ \mathrm{m}\mathit{\left(}{\mathrm{N}}_{\mathrm{a}}{\mathrm{CO}}_3\right)={10}^{10.00-10.329}.\frac{5\mathrm{g}}{84.01\mathrm{g}/{\mathrm{mol}}^2}.105.99\mathrm{g}/{\mathrm{mol}}^{2.96\mathrm{g}} \\ \mathrm{m}=2.96\mathrm{g} \end{gathered}$$