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.
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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