How many milliliters of 0.120 M NaOH are required to titrate 50.0 mL of 0.0998 M butanoic acid to the equivalence point? 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

First, we need to write the balanced chemical equation for the reaction between NaOH and butanoic acid:
CH3CH2CH2COOH + NaOH → CH3CH2CH2COONa + H2O

From the equation, we can see that the reaction is a 1:1 stoichiometry between butanoic acid and NaOH. This means that the number of moles of NaOH required to reach the equivalence point is equal to the number of moles of butanoic acid present in the 50.0 mL solution.

We can calculate the number of moles of butanoic acid present in the solution using the formula:

moles of acid = concentration × volume
moles of acid = 0.0998 mol/L × 0.0500 L
moles of acid = 0.00499 mol

Since the reaction is 1:1 stoichiometry, we know that we also need 0.00499 mol of NaOH to reach the equivalence point.

To calculate the volume of NaOH solution required, we can use the formula:

moles of NaOH = concentration × volume
0.00499 mol = 0.120 mol/L × volume
volume = 0.0416 L

Converting to milliliters:

volume = 41.6 mL

Therefore, the answer is C) 41.6 mL.

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