Question

If it takes 5 ml of 1.4 m naoh to neutralize 150 ml of hcl with an unknown concentration, what was the original concentration of the acid?

Answer 1

Answer:

0.04667 M was the original concentration of the acid.

Explanation:

To calculate the concentration of acid, we use the equation given y 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:

$n_1=1\\M_1=?M\\V_1=150 mL\\n_2=1\\M_2=1.4 M\\V_2=5 mL$

Putting values in above equation, we get:

$1\times M_1\times 150 mL=1\times 1.4 M\times 5 mL\\\\M_1=0.04667 M$

0.04667 M was the original concentration of the acid.