Question

Find how many milliliters of NaOH should be used to reach the half-equivalence point during the titration of 20.00 mL 0.888 M butanoic acid, CH3CH2CH2COOH (Ka = 1.54×10–5) with 0.425 M NaOH solution. Enter 2 decimal places.

Answer 1

Answer: The volume of NaOH solution required to reach the half-equivalence point is 0.09 mL

Explanation:

The chemical equation for the dissociation of butanoic acid follows:

$CH_3CH_2CH_2COOH\rightleftharpoons CH_3CH_2CH_2COO^-+H^+$

The expression of $K_a$ for above equation follows:

$K_a=\frac{[CH_3CH_2CH_2COO^-][H^+]}{[CH_3CH_2CH_2COOH]}$

We are given:

$[CH_3CH_2CH_2COOH]=0.888M\\K_a=1.54\times 10^{-5}$

$[CH_3CH_2CH_2COO^-]=[H^+]$

Putting values in above expression, we get:

$1.54\times 10^{-5}=\frac{[H^+]^2}{0.888}$

$[H^+]=-0.0037,0.0037$

Neglecting the negative value because concentration cannot be negative

To calculate the volume of base, we use the equation given by neutralization reaction:

$n_1M_1V_1=n_2M_2V_2$

where,

$n_1,M_1\text{ and }V_1$ are the n-factor, molarity and volume of acid which is butanoic acid

$n_2,M_2\text{ and }V_2$ are the n-factor, molarity and volume of base which is NaOH.

We are given:

$n_1=1\\M_1=\frac{0.0037}{2}M\text{ (half equivalence)}\\V_1=20.00mL\\n_2=1\\M_2=0.425M\\V_2=?mL$

Putting values in above equation, we get:

$1\times \frac{0.0037}{2}\times 20.00=1\times 0.425\times V_2\\\\V_2=\frac{1\times 0.0037\times 20}{1\times 0.425\times 2}=0.087mL$

Hence, the volume of NaOH solution required to reach the half-equivalence point is 0.09 mL