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

Step-by-step 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* M_1* 10.4=1* 0.138* 10.6\\\\M_1=(1* 0.138* 10.6)/(1* 10.4)=0.141M`

Hence, the molarity of hydroiodic acid solution is 0.141 M