Question

7. A student titrated a 15.00-mL sample of a solution containing a weak, monoprotic acid with NaOH. The titration required 17.73 mL of 0.1036 M NaOH to reach the equivalence point. Calculate the concentration (in M) of the weak acid in the sample.

Answer 1

Answer:  The concentration of the weak acid in the sample is 0.1224 M

Explanation:

The balanced chemical reaction is :

$HA+NaOH\rightarrow NaA+H_2O$

To calculate the volume 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 $HA$

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

We are given:

$n_1=1\\M_1=?\\V_1=15.00mL\\n_2=1\\M_2=0.1036M\\V_2=17.73mL$

Putting values in above equation, we get:

$1\times M_1\times 15.00=1\times 0.1036\times 17.73\\\\M_1=0.1224$

Thus the concentration (in M) of the weak acid in the sample is 0.1224