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 → Na2CO3 + 2 H2O


The concentration of the carbonic acid is _____.

It Takes 60 ML Of 0.20 M Of Sodium Hydroxide (NaOH) To Neutralize 25 ML Of Carbonic Acid (H2CO3) For

Answers

Answer 1

It requires 60 mL of 0.20 M sodium hydroxide [tex](NaOH)[/tex] to neutralize 25 mL of carbonic acid [tex](H_2CO_3)[/tex], hence the carbonate ions concentration is [tex]0.24M[/tex].

Given:

Reaction:

[tex]\to \bold{2NaOH + H_2CO_3 \to Na_2CO_3 + 2H_20 }[/tex]

[tex]NaOH[/tex] volume [tex](V_B) = 60 \ ml[/tex]  

[tex]H_2CO_3[/tex] Volume [tex](V_A) = 25\ ml[/tex]  

[tex]NaOH[/tex] Molarity [tex](C_B) = 0.20\ M[/tex]

[tex]H_2CO_3[/tex] moles [tex](n_A) = 1[/tex]

[tex]NaOH[/tex] moles [tex](n_B) = 2[/tex]

To find:

[tex]H_2CO_3[/tex] Molarity [tex](C_A) =?[/tex]

Solution:

Using the neutralization reaction:  

[tex]\to \frac{C_AV_A}{C_BV_B} =\frac{n_A}{n_B} \\\\[/tex]

[tex]\to C_B = 0.2\ M \\\\ \to n_A = 1 \\\\ \to n_B = 2 \\\\ \to V_B = 60\ ml \\\\ \to V_A = 25\ ml[/tex]

Calculating the [tex]C_A[/tex]:

 [tex]\to \frac{C_A \times 25}{0.2 \times 60} =\frac{1}{2} \\\\\to C_A =\frac{1 \times 0.2 \times 60}{2 \times 25} \\\\\to C_A =\frac{1 \times 2 \times 12}{2 \times 5 \times 10} \\\\ \to C_A =\frac{ 12}{ 5 \times 10} \\\\\to C_A = \frac{6}{5 \times 5} \\\\\to C_A = \frac{6}{25} \\\\\to C_A=0.24\ M[/tex]

Therefore, the concentration of carbonic acid is "0.24M".

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Answer 2

The concentration of carbonic acid( H₂CO₃) is 0.24 M.

The given reaction:

2 NaOH + H₂CO₃ → Na₂CO₃ + 2H₂O

Volume of NaOH  = 60mL

Volume of H₂CO₃ = 25 mL

Molarity of NaOH= 0.20M

To find:

Molarity of H₂CO₃=?

2 moles of NaOH and 1 mol of H₂CO₃ reacts to give 1 mol of Na₂CO₃.

Using the neutralization reaction:  

Consider A to be NaOH and B to be  H₂CO₃

[tex]\frac{\text{Number of moles of A}}{\text{Number of moles of B}} =\frac{\text{Molarity of A*Volume of A}}{\text{Molarity of B*Volume of B}}[/tex]

On substituting the values in order to calculate molarity of H₂CO₃:

[tex]\text{Molarity of B}=\frac{1*0.2*60}{2*25} \\\\\text{Molarity of B}=0.24M[/tex]

Therefore, the concentration of H₂CO₃ is 0.24 M.

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