If a 25.0 mL solution of acetic acid of unknown concentration required 30.0 mL of 0.100M...
If a 25.0 mL solution of acetic acid of unknown concentration required 30.0 mL of 0.100M NaOH solution to be neutralized, what is the concentration of the unknown substance.
Homework Answers
Balanced chemical equation is:
NaOH + CH3COOH ---> CH3COONa + H2O
Here:
M(NaOH)=0.1 M
V(NaOH)=30.0 mL
V(CH3COOH)=25.0 mL
According to balanced reaction:
1*number of mol of NaOH =1*number of mol of CH3COOH
1*M(NaOH)*V(NaOH) =1*M(CH3COOH)*V(CH3COOH)
1*0.1*30.0 = 1*M(CH3COOH)*25.0
M(CH3COOH) = 0.12 M
Answer: 0.120 M
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