Determine the volume in mL of 0.22 M NaOH(aq) needed to reach the equivalence (stoichiometric) point in the titration of 38 mL of 0.24 M propanoic acid(aq).

Question

Determine the volume in mL of 0.22 M NaOH(aq) needed to reach the equivalence (stoichiometric) point in the titration of 38 mL of 0.24 M propanoic acid(aq). 

The K_a of propanoic acid is 1.3 x 10^-5.

the answer is 41.45

please show your steps, thank you :)

Answer 1

Propanoic acid = CH3CH2COOH

Neutralization reaction with NaOH is ...

CH3CH2COOH + NaOH ===> CH3CH2COONa + H2O

moles CH3CH2COOH present = 0.038 L x 0.24 mole/L = 0.00912 moles

moles NaOH required = 0.00912 moles

Volume NaOH required: (x L)(0.22 mol/L) = 0.00912 moles

x = 0.04145 L = 41.45 mls

(NOTE: the Ka for propionic acid is not required to answer this question. It would be needed if one wanted to find the pH of the resulting solution).