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Carbonic acid, H2CO3, has two acidic hydrogens. A solution containing an unknown concentration of carbonic acid is titrated with

potassium hydroxide. It requires 20.0 mL of 3.840 M KOH solution to titrate both acidic protons in 50.2 mL of the carbonic acid solution.
1. Write a balanced net ionic equation for the neutralization reaction. Include physical states.
2. Calculate the molarity of the carbonic acid solution.

1 answer:

KiRa [2.9K]1 month ago

6 0

Answer:

1) The net ionic equation:

$2H^+(aq)+2OH^-(aq)\rightarrow 2H_2O(l)$

2) The molarity of the carbonic acid solution equals 0.765 M.

Explanation:

1) In aqueous carbonic acid, both carbonate ions and hydrogen ions exist:

$H_2CO_3(aq)\rightarrow 2H^+(aq)+CO_3^{2-}(aq)$ ..[1]

In potassium hydroxide solution, potassium ions and hydroxide ions are found:

$KOH(aq)\rightarrow K^+(aq)+OH^{-}(aq)$ ..[2]

In potassium carbonate solution, potassium ions and carbonate ions are present:

$K_2CO_3(aq)\rightarrow 2K^+(aq)+CO_3^{2-}(aq)$ ..[3]

$H_2CO_3(aq)+2KOH(aq)\rightarrow K_2CO_3(aq)+2H_2O(l)$

From these:[1], [2], and [3]:

$2H^+(aq)+CO_3^{2-}(aq)+2K^+(aq)+2OH^{-}(aq)\rightarrow 2K^+(aq)+CO_3^{2-}(aq)+H_2O(l)$

To formulate the net ionic equation, eliminate common ions on both sides:

$2H^+(aq)+2OH^-(aq)\rightarrow 2H_2O(l)$

2)

For determining acid concentration, utilize the equation from the neutralization reaction:

$n_1M_1V_1=n_2M_2V_2$

In which,

$n_1,M_1\text{ and }V_1$ indicates the n-factor, molarity and volume related to acid represented as $H_2CO_3$ .

$n_2,M_2\text{ and }V_2$ indicates n-factor, molarity and volume concerning the KOH base.

Provided information includes:

$n_1=2\\M_1=?\\V_1=50.0 mL\\n_2=1\\M_2=3.840 M\\V_2=20.0 mL$

By plugging in these values to the equation, we find:

$M_1=\frac{1\times 3.840 M\times 20.0 mL}{2\times 50.2 mL}=0.765 M$

The molarity of the carbonic acid solution is found to be 0.765 M.

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