How many millilitres of 0.120 mol L^-1 NaOH are required to titrate 50.0 mL of 0.0998 mol L^-1 butanoic acid to the equivalence point? Butanoic acid is monoprotic. The K_a of butanoic acid is 1.5 × 10^-5. A)4.90 B)50.0 C)41.6 D)60.1 E)4.65

Question

Quizplus · Accepted Answer

The balanced chemical equation for the reaction between NaOH and butanoic acid is:
NaOH + HC4H7O2 → NaC4H7O2 + H2O

From the equation, we can see that 1 mole of NaOH reacts with 1 mole of HC4H7O2. Therefore, the amount of NaOH required to react with 0.0998 mol HC4H7O2 is also 0.0998 mol.

At the equivalence point, the moles of NaOH should be equal to the moles of HC4H7O2.

n(NaOH) = n(HC4H7O2)
0.0998 mol = (C_vol/L_vol) x 0.120 mol
C_vol/L_vol = 0.8317
L_vol = 0.0500 L (given)

Therefore, the volume of NaOH required to react with 50.0 mL of 0.0998 mol/L HC4H7O2 is:

V(NaOH) = (0.8317 mol/L) x 0.0500 L = 0.0416 L or 41.6 mL

The answer is therefore (C) 41.6.

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