Question

If 23.9 mL of a NaOH solution is required to titrate 25.00mL of a 0.105M CH3CO2H solution, what is the concentration of the NaOH?

Answer 1

The balanced equation is

CH3COOH + NaOH ------> CH3COONa + H2O

Number of moles of CH3CO2H = molarity * volume of solution in L

Number of moles of CH3CO2H = 0.105 * 0.025 = 0.00263 mole

From the balanced equation we can say that

1 mole of CH3COOH requires 1 mole of NaOH so

0.00263 mole of CH3COOH will require

= 0.00263 mole of CH3COOH *(1 mole of NaOH / 1 mole of CH3COOH)

= 0.00263 mole of NaOH

Molarity = number of moles of NaOH / volume of solution in L

Molarity = 0.00263 / 0.0239 = 0.110 M

Therefore, the concentration of NaOH = 0.110 M