Question

What volume of 0.235 M H2SO4 is needed to titrate 40.0 mL of 0.0500 M Na2CO M NaCO3 solution?

Answer 1

Answer: The volume of 0.235 M $H_2SO_4$ needed to titrate 40.0 mL of 0.0500 M $Na_2CO_3$ is 8.51 ml

Explanation:

To calculate the volume 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 $H_2SO_4$

$n_2,M_2\text{ and }V_2$ are the n-factor, molarity and volume of base which is $Na_2CO_3$

We are given:

$n_1=2\\M_1=0.235M\\V_1=?mL\\n_2=2\\M_2=0.0500M\\V_2=40.0mL$

Putting values in above equation, we get:

$2\times 0.235\times V_1=2\times 0.0500\times 40.0\\\\V_1=8.51mL$

Thus volume of 0.235 M $H_2SO_4$ needed to titrate 40.0 mL of 0.0500 M $Na_2CO_3$ is 8.51 ml