Question

A student titrated a 25.00-mL sample of a solution containing an unknown weak, diprotic acid (H2A) with N2OH. If the titration required 17.73 mL of 0.1036 M N2OH to completely neutralize the acid, calculate the concentration (in M) of the weak acid in the sample.
(a) 9.184 x 10 M
(b) 3.674 x 10-2 M
(c) 7.304 x 10-2 M
(d) 7.347 x 10-2 M
(e) 1.469 x 101 M

Answer 1

Answer: The concentration of weak acid is $3.674\times 10^{-2}M$

Explanation:

To calculate the concentration of acid, 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 $H_2A$

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

We are given:

$n_1=2\\M_1=?M\\V_1=25.00mL\\n_2=1\\M_2=0.1036M\\V_2=17.73mL$

Putting values in above equation, we get:

$2\times M_1\times 25.00=1\times 0.1036\times 17.73\\\\M_1=\frac{1\times 0.1036\times 17.73}{2\times 25.00}=3.674\times 10^{-2}M$

Hence, the concentration of weak acid is $3.674\times 10^{-2}M$