It takes 60 mL of 0.20 M of sodium hydroxide (NaOH) to neutralize 25 mL of carbonic acid (H2CO3) for the following chemical reaction:

2 NaOH + H2CO3 → N2CO3 + 2 NaOH

The concentration of the carbonic acid is _____.
A) 0.48m
B) 0.10m
C) 0.96m
d)0.24m

Answers

Answer 1

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

Learn more here: https://brainly.com/question/25943090


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Answer:

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Answers

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