Đáp án:

a, 

$$NaOH+CH_3COONa\to CH_3COONa+H_2O$$ 

$$2NaOH+(COOH)_2\to (COONa)_2+2H_2O$$ 

b,

$$2.V_{(COOH)_2}.N_{(COOH)_2}=V_{NaOH}.N_{NaOH}$$ 

$$\Rightarrow 2.0,1.10,00=11,50.N_{NaOH}$$ 

$$\Rightarrow N_{NaOH}=0,174N$$ 

$$V_{NaOH}.N_{NaOH}=V_{CH_3COOH}.N_{CH_3COOH}$$ 

$$\Rightarrow 0,174.12,60=10,00.N_{CH_3COOH}$$ 

$$\Rightarrow N_{CH_3COOH}=0,219N$$ 

c,

$$C_{M_{CH_3COOH}}=0,219M$$ 

$$\Rightarrow n_{CH_3COOH}=0,219.10,00.10^{-3}=2,19.10^{-3}$$ mol 

Vậy $$C\%(g/L)=\dfrac{2,19.10^{-3}.60}{10.10^{-3}}=13,14$$