Respuesta :
Answer:
1) Net ionic equation :
[tex]2H^+(aq)+2OH^-(aq)\rightarrow 2H_2O(l)[/tex]
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.:
[tex]H_2CO_3(aq)\rightarrow 2H^+(aq)+CO_3^{2-}(aq)[/tex] ..[1]
In aqueous potassium hydroxide , potassium ions and hydroxide ion is present.:
[tex]KOH(aq)\rightarrow K^+(aq)+OH^{-}(aq)[/tex] ..[2]
In aqueous potassium carbonate , potassium ions and carbonate ion is present.:
[tex]K_2CO_3(aq)\rightarrow 2K^+(aq)+CO_3^{2-}(aq)[/tex] ..[3]
[tex]H_2CO_3(aq)+2KOH(aq)\rightarrow K_2CO_3(aq)+2H_2O(l)[/tex]
From one:[1] ,[2] and [3]:
[tex]2H^+(aq)+CO_3^{2-}(aq)+2K^+(aq)+2OH^{-}(aq)\rightarrow 2K^+(aq)+CO_3^{2-}(aq)+H_2O(l)[/tex]
Cancelling common ions on both sides to get net ionic equation :
[tex]2H^+(aq)+2OH^-(aq)\rightarrow 2H_2O(l)[/tex]
2)
To calculate the concentration of acid, we use the equation given by neutralization reaction:
[tex]n_1M_1V_1=n_2M_2V_2[/tex]
where,
[tex]n_1,M_1\text{ and }V_1[/tex] are the n-factor, molarity and volume of acid which is [tex]H_2CO_3[/tex]
[tex]n_2,M_2\text{ and }V_2[/tex] are the n-factor, molarity and volume of base which is KOH.
We are given:
[tex]n_1=2\\M_1=?\\V_1=50.0 mL\\n_2=1\\M_2=3.840 M\\V_2=20.0 mL[/tex]
Putting values in above equation, we get:
[tex]M_1=\frac{1\times 3.840 M\times 20.0 mL}{2\times 50.2 mL}=0.765 M[/tex]
0.765 M is the molarity of the carbonic acid solution.
