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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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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