Question

Suppose you have a 0.0423 L sample of KOH of unknown concentration. If this amount of KOH is used to exactly titrate 0.0350 L of HCl that has a concentration of 0.225 M, what is the molarity of the KOH? HINT: Remember to first get your balanced equation.

Answer 1

0.186 M is the molarity of the KOH.

Step-by-step explanation:

`HCl(aq)+KOH(aq)\rightarrow KCl(aq)+H_2O(aq)`

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
`HCl`

`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=0.225 M\\V_1=0.0350 L\\n_2=1\\M_2=?\\V_2=0.0423 L`

Putting values in above equation, we get:

`M_2=(n_1M_1V_1)/(n_2V_2)=(1* 0.225 M* 0.0350 L)/(1* 0.0423L)`

`M_2=0.186 M`

0.186 M is the molarity of the KOH.