Question

A sample of solid calcium hdroxide, Ca(OH)2 is allowed to stand in water until a saturated solution is formed. A titration of 75.00mL of this solution with 5.00 x 10-2 M HCl 36.6 mL of the acid to reach the end point
Ca(OH)2 + 2HCl ? CaCl + 2H2O
What is the molarity?

Answer 1

Answer: The concentration of $Ca(OH)_2$ is 0.0122 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 $HCl$

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

We are given:

$n_1=1\\M_1=5.00\times 10^{-2}M=0.05M\\V_1=36.6mL\\n_2=2\\M_2=?M\\V_2=75mL$

Putting values in above equation, we get:

$1\times 0.05\times 36.6=2\times M_2\times 75\\\\M_2=0.0122M$

Hence, the concentration of $Ca(OH)_2$ is 0.0122 M.