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lawyer [7]
3 years ago
9

PLEASEEEE HELP MEE!

Chemistry
1 answer:
Ugo [173]3 years ago
7 0

Answer:

2nd option

Explanation:

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Answer:

1) Net ionic equation :

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

2) 0.765 M is  the molarity of the carbonic acid solution.

Explanation:

1) In aqueous carbonic acid , carbonate ions and hydrogen ion is present.:

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

In aqueous potassium hydroxide , potassium ions and hydroxide ion is present.:

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

In aqueous potassium carbonate , potassium ions and carbonate ion is 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 one:[1] ,[2] and [3]:

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

Cancelling common ions on both sides to get net ionic equation :

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

2)

To calculate the concentration of acid, we use the equation given by neutralization reaction:

n_1M_1V_1=n_2M_2V_2

where,

n_1,M_1\text{ and }V_1 are the n-factor, molarity and volume of acid which is H_2CO_3

n_2,M_2\text{ and }V_2 are the n-factor, molarity and volume of base which is KOH.

We are given:

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

Putting values in above equation, we get:

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

0.765 M is  the molarity of the carbonic acid solution.

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