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.

Based on this data, calculate the concentration of the barium hydroxide solution.

[Ba(OH)2] = ___ M

Answer 1

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

Step-by-step 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* 0.1023* 12.58=2* M_2* 10.00\\\\M_2=0.129M`

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