Question

An aqueous solution of hydroiodic acid is standardized by titration with a 0.138 M solution of potassium hydroxide.If 10.6 mL of base are required to neutralize 10.4 mL of the acid, what is the molarity of the hydroiodic acid solution?

Answer 1

Answer: The molarity of hydroiodic acid solution is 0.141 M

Explanation:

To calculate the concentration 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 KOH.

We are given:

$n_1=1\\M_1=?M\\V_1=10.4mL\\n_2=1\\M_2=0.138M\\V_2=10.6mL$

Putting values in above equation, we get:

$1\times M_1\times 10.4=1\times 0.138\times 10.6\\\\M_1=\frac{1\times 0.138\times 10.6}{1\times 10.4}=0.141M$

Hence, the molarity of hydroiodic acid solution is 0.141 M