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

Answer:

0.186 M is the molarity of the KOH.

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:

$M_2=\frac{n_1M_1V_1}{n_2V_2}=\frac{1\times 0.225 M\times 0.0350 L}{1\times 0.0423L}$

$M_2=0.186 M$

0.186 M is the molarity of the KOH.