Answer:
0.02685 M is the molarity of the barium hydroxide solution.
Explanation:
To calculate the concentration of acid, 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]HCl[/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] mm.
We are given:
[tex]n_1=11\\M_1=0.110 M\\V_1=14.4 mL\\n_2=2\\M_2=?\\V_2=29.5 mL[/tex]
Putting values in above equation, we get:
[tex]1\times 0.11 M\times 14.4 mL=2\times M_2\times 29.5 mL[/tex]
[tex]M_2=\frac{1\times 0.11 M\times 14.4 mL}{2\times 29.5 mL}=0.02685 M[/tex]
0.02685 M is the molarity of the barium hydroxide solution.
