Question

In a titration experiment, 31.4 mL of 1.120 M HCOOH is neutralized by 16.3 mL of Ba(OH)2. What is the concentration of the Ba(OH)2 solution?

Answer 1

Answer : The concentration of the $Ba(OH)_2$ solution is, 2.16 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 $HCOOH$

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

We are given:

$n_1=1\\M_1=1.120M\\V_1=31.4mL\\n_2=2\\M_2=?\\V_2=16.3mL$

Putting values in above equation, we get:

$1\times 1.120M\times 31.4mL=2\times M_2\times 16.3mL\\\\M_2=2.16M$

Thus, the concentration of the $Ba(OH)_2$ solution is, 2.16 M