Question

How would you make $ 100.0 \cdot ml$ of a $ 1.00 \cdot mol/L$ buffer solution with a pH of $ 10.80$ to be made using only sodium carbonate, sodium hydrogen carbonate and water? How much sodium carbonate and sodium hydrogen carbonate would you use?

Answer 1

Hint: for the given problem first we need to find out the concentration ratios, then we need to find out the concentrations and finally we will find out the amounts of reagents used to make the buffer solution.

Complete answer:
A buffer solution is the mixture of weak acid or weak base with its conjugate base or acid. It resists the small changes in the pH of the solution, i.e. if a small amount of acid or base is added to it, no pH change will be observed.
Calculation of concentration ratios –
The chemical equation for the equilibrium is
$ \eqalign{
  & HCO_3^ - + {H_2}O \rightleftharpoons CO_3^{2 - } + {H_3}{O^ + };\;{K_a}\; = 4.8 \times {10^{ - 11}};\;p{K_a}\; = 10.32 \cr
  & HA + {H_2}O \rightleftharpoons {A^ - } + {H_3}{O^ + } \cr} $
Now, for this reaction, the Henderson-Hasselbalch equation is
$ \eqalign{
  & pH\; = p{K_a} + log\left( {\dfrac{{[{A^ - }]}}{{[HA]}}} \right) \cr
  & 10.80 = 10.32{\text{ }} + \;log\left( {\dfrac{{[{A^ - }]}}{{[HA]}}} \right) \cr
  & log\left( {\dfrac{{[{A^ - }]}}{{[HA]}}} \right) = 10.80-10.32 = 0.48 \cr
  & \left( {\dfrac{{[{A^ - }]}}{{[HA]}}} \right) = {10^{0.48}}\; = 3.02\; \cr} $
Now, this indicates that $ pH > p{K_a}$ , so there should be more $ {A^ - }$ than $ HA$
Calculation of concentrations-
$ [{A^ - }] = 3.02[HA]$ , Also,
$ \eqalign{
  & [{A^ - }] + [HA] = 1.00mol/L \cr
  & 3.02[HA] + [HA] = 4.02[HA] = 1.00mol/L \cr
  & [HA] = \dfrac{{1.00mol/L}}{{4.02}} = 0.2488mol/L \cr
  & [{A^ - }] = 3.02[HA] = 3.02 \times 0.2488mol/L = 0.7512mol/L \cr} $
Calculate the masses of $NaHC{O_3}$ and of $ N{a_2}C{O_3}$
 Mass of $NaHC{O_3}$ can be calculated as
$Mas{s_{NaHC{O_3}}} = 0.1000L \times \dfrac{{0.2488 \cdot mol \cdot NaHC{O_3}}}{{1L}} \times \dfrac{{84.01 \cdot g \cdot NaHC{O_3}}}{{1 \cdot mol \cdot NaHC{O_3}}} = 2.09g$
Mass of $N{a_2}C{O_3}$ can be calculated as
$Mas{s_{N{a_2}C{O_3}}} = 0.1000L \times \dfrac{{0.7512 \cdot mol \cdot N{a_2}C{O_3}}}{{1L}} \times \dfrac{{106.0 \cdot g \cdot N{a_2}C{O_3}}}{{1 \cdot mol \cdot N{a_2}C{O_3}}} = 7.96g$
So, for making $ 100.0 \cdot ml$ of a $ 1.00 \cdot mol/L$ buffer solution with a pH of $ 10.80$ to be made using only sodium carbonate, sodium hydrogen carbonate and water, we require $2.09gram$ of $NaHC{O_3}$ and $2.09gram$ of $N{a_2}C{O_3}$ and both are added to water in a volumetric flask of capacity $100ml$ and the volume is made up to $100ml$ using distilled water.

Note:
Calculation should be done carefully. The concentration ratio helps in the calculation of concentration and similarly mass can be calculated using concentration. The ${K_a}\& p{K_a}$ values are taken from the standard tables.