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

Hint: 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\[{{\text{C}}_{\text{6}}}{{\text{H}}_{\text{5}}}{\text{COOH}}\].