Question

A 0.2g sample of benzoic acid ${{\text{C}}_{\text{6}}}{{\text{H}}_{\text{5}}}{\text{COOH}}$ is titrated with a 0.120M ${\text{Ba(OH}}{{\text{)}}_{\text{2}}}$solution. What volume of the ${\text{Ba(OH}}{{\text{)}}_{\text{2}}}$solution is required to reach the equivalence point?
Molar mass of $C_6H_5COOH$ = $122.1\, gmol^{-1}$
A. 6.82 ml
B. 13.6 ml
C. 17.6 ml
D. 35.2 ml

Answer 1

Write the balanced acid-base reaction between ${{\text{C}}_{\text{6}}}{{\text{H}}_{\text{5}}}{\text{COOH}}$ and ${\text{Ba(OH}}{{\text{)}}_{\text{2}}}$. Calculate the moles of benzoic acid and then using the stoichiometric ratio, calculate the moles of ${\text{Ba(OH}}{{\text{)}}_{\text{2}}}$. Finally using the moles and molar concentration of ${\text{Ba(OH}}{{\text{)}}_{\text{2}}}$ , calculate the volume of it.

Formula Used:
${\text{Moles = }}\dfrac{{{\text{mass}}}}{{{\text{Molar mass}}}}$
${\text{Molarity = }}\dfrac{{{\text{ moles }}}}{{{\text{ L of solution}}}}$

Complete step by step answer:
The balanced acid-base reaction between ${{\text{C}}_{\text{6}}}{{\text{H}}_{\text{5}}}{\text{COOH}}$ and ${\text{Ba(OH}}{{\text{)}}_{\text{2}}}$ is as follows:
${\text{2}}{{\text{C}}_{\text{6}}}{{\text{H}}_{\text{5}}}{\text{COOH (aq) + Ba(OH}}{{\text{)}}_{\text{2}}}{\text{(aq) }} \rightleftharpoons {\text{ Ba(}}{{\text{C}}_{\text{6}}}{{\text{H}}_{\text{5}}}{\text{COO}}{)_2}({\text{aq) + 2}}{{\text{H}}_{\text{2}}}{\text{O(l)}}$
Now, using the mass and molar mass of ${{\text{C}}_{\text{6}}}{{\text{H}}_{\text{5}}}{\text{COOH}}$ given to us calculate the moles of ${{\text{C}}_{\text{6}}}{{\text{H}}_{\text{5}}}{\text{COOH}}$.
${\text{Moles = }}\dfrac{{{\text{mass}}}}{{{\text{Molar mass}}}}$
Substitute 0.2g for the mass of ${{\text{C}}_{\text{6}}}{{\text{H}}_{\text{5}}}{\text{COOH}}$ and $122.1{\text{ gm}}{{\text{ol}}^{{\text{ - 1}}}}$ for molar mass of ${{\text{C}}_{\text{6}}}{{\text{H}}_{\text{5}}}{\text{COOH}}$ and calculate the moles of ${{\text{C}}_{\text{6}}}{{\text{H}}_{\text{5}}}{\text{COOH}}$ as follows:
${\text{Moles = }}\dfrac{{0.2{\text{ g}}}}{{122.1{\text{ gm}}{{\text{ol}}^{{\text{ - 1}}}}}} = 0.0{\text{0163 mol}}$
Now, using these moles of ${{\text{C}}_{\text{6}}}{{\text{H}}_{\text{5}}}{\text{COOH}}$ and balanced chemical reaction calculate the moles of ${\text{Ba(OH}}{{\text{)}}_{\text{2}}}$ at as follows:
From the balanced reaction, we can say that 2 moles of ${{\text{C}}_{\text{6}}}{{\text{H}}_{\text{5}}}{\text{COOH}}$ reacts with 1 mole of ${\text{Ba(OH}}{{\text{)}}_{\text{2}}}$.
So, $0.0{\text{0163 mol }}{{\text{C}}_{\text{6}}}{{\text{H}}_{\text{5}}}{\text{COOH}} \times \dfrac{{1{\text{ mol Ba(OH}}{{\text{)}}_{\text{2}}}{\text{ }}}}{{2{\text{ mol }}{{\text{C}}_{\text{6}}}{{\text{H}}_{\text{5}}}{\text{COOH }}}} = 8.19 \times {10^{{\text{ - 4}}}}{\text{ mol Ba(OH}}{{\text{)}}_{\text{2}}}$
Now, we have moles of ${\text{Ba(OH}}{{\text{)}}_{\text{2}}}$ and also we have given molar concentration of ${\text{Ba(OH}}{{\text{)}}_{\text{2}}}$.
Hence, calculate the volume of ${\text{Ba(OH}}{{\text{)}}_{\text{2}}}$require to reach the equivalence point as follows:
${\text{Molarity = }}\dfrac{{{\text{ moles }}}}{{{\text{ L of solution}}}}$
Substitute $8.19 \times {10^{{\text{ - 4}}}}{\text{ mol Ba(OH}}{{\text{)}}_{\text{2}}}$ and 0.120M ${\text{Ba(OH}}{{\text{)}}_{\text{2}}}$ in the molarity equation and calculate the volume of ${\text{Ba(OH}}{{\text{)}}_{\text{2}}}$require to reach the equivalence point.
${\text{Litres of Ba(OH}}{{\text{)}}_{\text{2}}}{\text{ solution}} = 0.00682{\text{ L}}$
Convert the volume of ${\text{Ba(OH}}{{\text{)}}_{\text{2}}}$ solution from L to ml.
1 L = 1000 ml
$0.00682{\text{ L}} \times \dfrac{{1000{\text{ ml}}}}{{1{\text{ L}}}} = 6.82{\text{ ml}}$
Thus, 6.82 ml of ${\text{Ba(OH}}{{\text{)}}_{\text{2}}}$ is required to reach the equivalence point.

**Hence, the correct option is (A) 6.82 ml

Note: **
Acid is a proton donor species and the base is a proton acceptor species. It is very important to write the correct balance reaction as a mole calculation depends on the stoichiometric ratio. From the balanced reaction, we can say that at the equivalence point 1 mole of ${\text{Ba(OH}}{{\text{)}}_{\text{2}}}$ reacts with 2 moles of${{\text{C}}_{\text{6}}}{{\text{H}}_{\text{5}}}{\text{COOH}}$.