Answer: The concentration of [tex]Ba(OH)_2[/tex] comes out to be 0.129 M.
Explanation:
To calculate the concentration of base, we use the equation given by neutralization reaction:
[tex]n_1M_1V_1=n_2M_2V_2[/tex]
where,
[tex]n_1,M_1\text{ and }V_1[/tex] are the n-factor, molarity and volume of acid which is [tex]H_2SO_4[/tex]
[tex]n_2,M_2\text{ and }V_2[/tex] are the n-factor, molarity and volume of base which is [tex]Ba(OH)_2[/tex]
We are given:
[tex]n_1=2\\M_1=0.1023M\\V_1=12.58mL\\n_2=2\\M_2=?M\\V_2=10.00mL[/tex]
Putting values in above equation, we get:
[tex]2\times 0.1023\times 12.58=2\times M_2\times 10.00\\\\M_2=0.129M[/tex]
Hence, the concentration of [tex]Ba(OH)_2[/tex] comes out to be 0.129 M.
