Question

An aqueous solution of hydroiodic acid is standardized by titration with a 0.186 M solution of calcium hydroxide. If 26.5 mL of base are required to neutralize 20.3 mL of the acid, what is the molarity of the hydroiodic acid solution? M hydroiodic acid

Answer 1

Answer: The molarity of hydroiodic acid in the titration is 0.485 M.

Explanation:

To calculate the molarity of acid, 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 $HI$

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

We are given:

$n_1=1\\M_1=?M\\V_1=20.3mL\\n_2=2\\M_2=0.186M\\V_2=26.5mL$

Putting values in above equation, we get:

$1\times M_1\times 20.3=2\times 0.186\times 26.5\\\\M_1=0.485M$

Hence, the molarity of hydroiodic acid is 0.485M.