Answer:
1) Net ionic equation :
![2H^+(aq)+2OH^-(aq)\rightarrow 2H_2O(l)](https://tex.z-dn.net/?f=2H%5E%2B%28aq%29%2B2OH%5E-%28aq%29%5Crightarrow%202H_2O%28l%29)
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.:
..[1]
In aqueous potassium hydroxide , potassium ions and hydroxide ion is present.:
..[2]
In aqueous potassium carbonate , potassium ions and carbonate ion is present.:
..[3]
![H_2CO_3(aq)+2KOH(aq)\rightarrow K_2CO_3(aq)+2H_2O(l)](https://tex.z-dn.net/?f=H_2CO_3%28aq%29%2B2KOH%28aq%29%5Crightarrow%20K_2CO_3%28aq%29%2B2H_2O%28l%29)
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)](https://tex.z-dn.net/?f=2H%5E%2B%28aq%29%2BCO_3%5E%7B2-%7D%28aq%29%2B2K%5E%2B%28aq%29%2B2OH%5E%7B-%7D%28aq%29%5Crightarrow%202K%5E%2B%28aq%29%2BCO_3%5E%7B2-%7D%28aq%29%2BH_2O%28l%29)
Cancelling common ions on both sides to get net ionic equation :
![2H^+(aq)+2OH^-(aq)\rightarrow 2H_2O(l)](https://tex.z-dn.net/?f=2H%5E%2B%28aq%29%2B2OH%5E-%28aq%29%5Crightarrow%202H_2O%28l%29)
2)
To calculate the concentration of acid, we use the equation given by neutralization reaction:
![n_1M_1V_1=n_2M_2V_2](https://tex.z-dn.net/?f=n_1M_1V_1%3Dn_2M_2V_2)
where,
are the n-factor, molarity and volume of acid which is ![H_2CO_3](https://tex.z-dn.net/?f=H_2CO_3)
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](https://tex.z-dn.net/?f=n_1%3D2%5C%5CM_1%3D%3F%5C%5CV_1%3D50.0%20mL%5C%5Cn_2%3D1%5C%5CM_2%3D3.840%20M%5C%5CV_2%3D20.0%20mL)
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](https://tex.z-dn.net/?f=M_1%3D%5Cfrac%7B1%5Ctimes%203.840%20M%5Ctimes%2020.0%20mL%7D%7B2%5Ctimes%2050.2%20mL%7D%3D0.765%20M)
0.765 M is the molarity of the carbonic acid solution.