Question

181 mL of an unknown HCl solution was neutralized in a titration with 36.2 mL of 0.250 M NaOH. What is the molarity of the unknown HCl solution

Answer 1

Answer: The molarity of the unknown HCl solution is 0.05 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 $HCl$

$n_2,M_2\text{ and }V_2$ are the n-factor, molarity and volume of base which is NaOH.

We are given:

$n_1=1\\M_1=?\\V_1=181mL\\n_2=1\\M_2=0.250M\\V_2=36.2mL$

Putting values in above equation, we get:

$1\times M_1\times 181=1\times 0.250\times 36.2\\\\M_1=0.05M$

Thus the molarity of the unknown HCl solution is 0.05 M