How many milliliters of 0.120 M NaOH are required to titrate 50.0 mL of 0.0998 M formic acid to the equivalence point? The K_a of formic acid is 1.8 × 10^-4. A) 3.82 mL B) 50.0 mL C) 41.6 mL D) 60.1 mL E) 3.57 mL

Question

Quizplus · Accepted Answer

To solve this problem, we can use the equation:

KS1U1B1aS1U1B0 + NaOH → H2O + NaS1U1B1aS1U1P0

From this equation, we can see that the 1 mole of NaOH reacts with 1 mole of formic acid (KS1U1B1aS1U1B0). Therefore, the number of moles of formic acid in 50.0 mL of 0.0998 M solution is:

mol of formic acid = M × volume in L = 0.0998 M × 0.0500 L = 4.99 × 0.001 = 0.00499 mol

The number of millimoles of formic acid is:

mmol of formic acid = 0.00499 mol × 1000 = 4.99 mmol

At the equivalence point, all the formic acid has reacted with NaOH, so the number of millimoles of NaOH used is also equal to 4.99 mmol.

Using the concentration of NaOH, we can calculate the volume of NaOH required:

mol of NaOH = M × volume in L

0.120 M × volume in L = 0.00499 mol

volume in L = 0.00499 mol ÷ 0.120 M = 0.0416 L

volume in mL = 41.6 mL

Therefore, the answer is C.

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