Question

Ranjit titrates a sample 10.00 mL of Ba(OH)2 solution to the endpoint using 12.58 mL of 0.1023 M H2SO4.

Answer 1

Answer: The concentration of $Ba(OH)_2$ comes out to be 0.129 M.

Explanation:

To calculate the concentration 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 $H_2SO_4$

$n_2,M_2\text{ and }V_2$ are the n-factor, molarity and volume of base which is $Ba(OH)_2$

We are given:

$n_1=2\\M_1=0.1023M\\V_1=12.58mL\\n_2=2\\M_2=?M\\V_2=10.00mL$

Putting values in above equation, we get:

$2\times 0.1023\times 12.58=2\times M_2\times 10.00\\\\M_2=0.129M$

Hence, the concentration of $Ba(OH)_2$ comes out to be 0.129 M.