Answer: The concentration of [tex]Ca(OH)_2[/tex] is 0.0122 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]HCl[/tex]
[tex]n_2,M_2\text{ and }V_2[/tex] are the n-factor, molarity and volume of base which is [tex]Ca(OH)_2[/tex]
We are given:
[tex]n_1=1\\M_1=5.00\times 10^{-2}M=0.05M\\V_1=36.6mL\\n_2=2\\M_2=?M\\V_2=75mL[/tex]
Putting values in above equation, we get:
[tex]1\times 0.05\times 36.6=2\times M_2\times 75\\\\M_2=0.0122M[/tex]
Hence, the concentration of [tex]Ca(OH)_2[/tex] is 0.0122 M.
