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

Question

Mashutka [201]

2 years ago

13

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

tion:
2 NaOH + H2CO3 → Na2CO3 + 2 H2O

The concentration of the carbonic acid is _____.

Chemistry

2 answers:

Rashid [163] · Answer 1 · 2023-04-17T19:58:17+03:00

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

Given:

Reaction:

$\to \bold{2NaOH + H_2CO_3 \to Na_2CO_3 + 2H_20 }$

$NaOH$ volume $(V_B) = 60 \ ml$

$H_2CO_3$ Volume $(V_A) = 25\ ml$

$NaOH$ Molarity $(C_B) = 0.20\ M$

$H_2CO_3$ moles $(n_A) = 1$

$NaOH$ moles $(n_B) = 2$

To find:

$H_2CO_3$ Molarity $(C_A) =?$

Solution:

Using the neutralization reaction:

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

$\to C_B = 0.2\ M \\\\ \to n_A = 1 \\\\ \to n_B = 2 \\\\ \to V_B = 60\ ml \\\\ \to V_A = 25\ ml$

Calculating the $C_A$ :

$\to \frac{C_A \times 25}{0.2 \times 60} =\frac{1}{2} \\\\\to C_A =\frac{1 \times 0.2 \times 60}{2 \times 25} \\\\\to C_A =\frac{1 \times 2 \times 12}{2 \times 5 \times 10} \\\\ \to C_A =\frac{ 12}{ 5 \times 10} \\\\\to C_A = \frac{6}{5 \times 5} \\\\\to C_A = \frac{6}{25} \\\\\to C_A=0.24\ M$