Answer: The molarity of calcium hydroxide in the solution is 0.1 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]HBr[/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=0.120M\\V_1=27.5mL\\n_2=2\\M_2=?M\\V_2=16.5mL[/tex]
Putting values in above equation, we get:
[tex]1\times 0.120\times 27.5=2\times M_2\times 16.5\\\\M_2=0.1M[/tex]
Hence, the molarity of [tex]Ca(OH)_2[/tex] in the solution is 0.1 M.
