Question

A $0.2g$ sample of benzoic acid, $C_6{H}_{5}COOH$, is tritated with a $0.120MBa(OH{)}_2$ solution. What volume of the $Ba(OH{)}_2$ solution is required to reach the equivalent point?
$\frac{Substance}{C_6{H}_{5}COOH}\frac{Molarmass}{122.1gmo{l}^{-1}}$

• A. $6.82mL$
• B. $13.6mL$
• C. $17.6mL$
• D. $35.2mL$

Answer 1

The correct option is A $6.82mL$

Given

Molar mass of Benzoic Acid $=122.1gmo{l}^{-1}$

Morality of $Ba(OH{)}_2=0.120mol/L$

$2E_6{H}_{5}COOH+Ba(OH{)}_2\to 2H_2+Ba(C_6{H}_{5}COO{)}_2$

From the given Molar mass, we find the number

of moles of Benzoic acid and Barium

hydrocide.

Moles of Benzoic acid $=\frac{GivenMass}{MolarMass}$

$=\frac{0.2}{122.1}=0.00163$ Moles

Moles of $Ba(OH{)}_2=\frac{0.00163}{2}=0.000815$

Now, Morality of $Ba(OH{)}_2=\frac{molesofsolute}{LitreofSolution}$

$=0.120M$ is required 1 Litre,

So $0.000815M$ is required in $\frac{1}{0.120}\times 0.000815$

$=0.00679$

$=6.8$