Question

0.450 L of 0.0500 M HCl is titrated to the equivalence point with 21.45 mL of a NaOH solution. What is the concentration (in M) of the NaOH solution that was added?

Answer 1

Answer:

1.0489 M is the concentration (in M) of the NaOH solution that was added

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 $HCl$

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

We are given:

$n_1=1\\M_1=0.0500 M\\V_1=0.450 L\\n_2=1\\M_2=?\\V_2=21.45 mL=0.02145 L$

Putting values in above equation, we get:

$1\times 0.0500 M\times 0.450 L=1\times M_2\times 0.02145 L\\\\M_2=1.0489 M$

1.0489 M is the concentration (in M) of the NaOH solution that was added