If It takes 60mL of 0.20M of sodium hydroxide (NaOH) to neutralize 25 mL of carbonic acid ( $H_2CO_3$ ), then the concentration of the carbonic acid is 0.24M

The reaction between NaOH solution and $H_2CO_3$ is written below

$2NaOH + H_2CO_3 \rightarrow Na_2CO_3 + 2H_2O$

Volume of NaOH, $V_B$ = 60 ml

Volume of $H_2CO_3$ , $V_A=25 ml$

Molarity of $H_2CO_3$ , $C_A=?$

Molarity of NaOH, $C_B=0.20M$

Number of moles of $H_2CO_3$ , $n_A=1$

Number of moles of NaOH, $n_B=2$

The mathematical equation for neutralization reaction is:

$\frac{C_AV_A}{C_BV_B} =\frac{n_A}{n_B}$

Substitute $C_B=0.2 M$ , $n_A=1$ , $n_B=2$ , $V_B$ = 60ml, and $V_A=25 ml$ into the equation above in order to solve for $C_A$

$\frac{C_A \times 25}{0.2 \times 60}=\frac{1}{2} \\\\50C_A=12\\\\C_B=\frac{12}{50} \\\\C_B=0.24M$

Therefore, the concentration of the carbonic acid is 0.24M

Learn more here: brainly.com/question/25943090

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