If It takes 60mL of 0.20M of sodium hydroxide (NaOH) to neutralize 25 mL of carbonic acid ([tex]H_2CO_3[/tex]), then the concentration of the carbonic acid is 0.24M
The reaction between NaOH solution and [tex]H_2CO_3[/tex] is written below
[tex]2NaOH + H_2CO_3 \rightarrow Na_2CO_3 + 2H_2O[/tex]
Volume of NaOH, [tex]V_B[/tex] = 60 ml
Volume of [tex]H_2CO_3[/tex], [tex]V_A=25 ml[/tex]
Molarity of [tex]H_2CO_3[/tex], [tex]C_A=?[/tex]
Molarity of NaOH, [tex]C_B=0.20M[/tex]
Number of moles of [tex]H_2CO_3[/tex], [tex]n_A=1[/tex]
Number of moles of NaOH, [tex]n_B=2[/tex]
The mathematical equation for neutralization reaction is:
[tex]\frac{C_AV_A}{C_BV_B} =\frac{n_A}{n_B}[/tex]
Substitute [tex]C_B=0.2 M[/tex], [tex]n_A=1[/tex], [tex]n_B=2[/tex], [tex]V_B[/tex] = 60ml, and [tex]V_A=25 ml[/tex] into the equation above in order to solve for [tex]C_A[/tex]
[tex]\frac{C_A \times 25}{0.2 \times 60}=\frac{1}{2} \\\\50C_A=12\\\\C_B=\frac{12}{50} \\\\C_B=0.24M[/tex]
Therefore, the concentration of the carbonic acid is 0.24M
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