How much solid \({\bf{NaC}}{{\bf{H}}_{\bf{3}}}{\bf{C}}{{\bf{O}}_{\bf{2}}} \bullet {\bf{3}}{{\bf{H}}_{\bf{2}}}{\bf{O}}\) must be added to \({\bf{0}}.{\bf{300}}{\rm{ }}{\bf{L}}\) of a \({\bf{0}}.{\bf{50}}{\rm{ }}{\bf{M}}\) acetic acid solution to give a buffer with a pH of 5.00? (Hint: Assume a negligible change in volume as the solid is added.)

This is a buffer solution because it is composed of a weak acid (acetic acid,\(C{H_3}COOH\)) and its salt\(\left( {NaC{H_3}CO{C_2} \cdot 3{H_2}O} \right)\). Therefore, the pH is calculated using the Henderson-Hasselbach equation

\(pH = p{K_a} + log\left( {\frac{{c(salt)}}{{c( acid )}}} \right)\)

Here we know the pH, we can calculate the \(p{K_a}\)using the \({K_a}\)value in the table in Appendix H in the book and we know the concentration of acetic acid. What we need to calculate is the mass of the \(NaC{H_3}C{O_2}H\)which we need to add. Since we know the volume of the solution, we can calculate the concentration of the salt (base) and from there the amount and the mass of the salt.

\(\begin{align}pH &= p{K_a} + log\left( {\frac{{c( salt )}}{{c( acid )}}} \right)\\log\left( {\frac{{c( salt )}}{{c( acid )}}} \right) &= pH - p{K_a}\\\frac{{c( salt )}}{{c( acid )}} &= 1{0^{pH - p{K_a}}}\\c( base ) = c( acid ) \times 1{0^{pH - p{K_a}}}\\p{K_a} &= - log{K_a}\\p{K_a} &= - log\left( {1.8 \times 1{0^{ - 5}}} \right)\\p{K_a} &= 4.74\\c( salt ) &= 0.50M \times 1{0^{5.00 - 4.74}}\\c( salt ) &= 0.91M\end{align}\)

Now that we know the concentration of the salt, we can calculate its amount using the formula:

\(\begin{align}n &= c \times V\\n &= 0.91\frac{{mol}}{L} \times 0.300L\\n &= 0.27\;mol\end{align}\)

Now we can calculate the mass of the salt using the formula:

\(m = M \cdot n\)

M is the molar mass, and it is calculated as the sum of all the relative atomic masses of the elements composing the molecule. We can look up the relative atomic masses in any periodic table of elements.

\(\begin{align}M\left( {{\rm{NaC}}{{\rm{H}}_3}{\rm{C}}{{\rm{O}}_2} \cdot 3{{\rm{H}}_2}{\rm{O}}} \right) &= ({\rm{Ar}}({\rm{Na}}) + 2{\rm{Ar}}({\rm{C}}) + 9{\rm{Ar}}({\rm{H}}) + 5{\rm{Ar}}({\rm{O}}))\frac{{\rm{g}}}{{{\rm{mol}}}}\\M\left( {{\rm{NaC}}{{\rm{H}}_3}{\rm{C}}{{\rm{O}}_2} \cdot 3{{\rm{H}}_2}{\rm{O}}} \right) &= (22.99 + 2 \cdot 12.01 + 9 \cdot 1.01 + 5 \cdot 15.99)\frac{{\rm{g}}}{{{\rm{mol}}}}\\{\rm{M}}\left( {{\rm{NaC}}{{\rm{H}}_3}{\rm{C}}{{\rm{O}}_2} \cdot 3{{\rm{H}}_2}{\rm{O}}} \right) &= 136.05\frac{{\rm{g}}}{{{\rm{mol}}}}\\m\left( {{\rm{NaCHCH}}{{\rm{O}}_3} \cdot 3{{\rm{H}}_2}{\rm{O}}} \right) &= 136.05\frac{{\rm{g}}}{{{\rm{mol}}}} \cdot 0.27\;{\rm{mol}}\\m\left( {{\rm{NaC}}{{\rm{H}}_3}{\rm{C}}{{\rm{O}}_2} \cdot 3{{\rm{H}}_2}{\rm{O}}} \right) &= 36.73\;{\rm{g}}\end{align}\)

To achieve a buffer solution of the required pH, we need to add\(36.73\;g\)of \(NaC{H_3}C{O_2} \cdot 3{H_2}O\)to the acetic acid solution.