What volume of 0.0500 M sodium hydroxide should be added to 250. mL of 0.100 M HCOOH? to obtain a solution with a pH of 4.50 [K_a(HCOOH) = 1.7 * 10^-4]

Question

Quizplus · Accepted Answer

First, we need to use the Ka expression for formic acid:

Ka = [H+][HCOO-]/[HCOOH]

And since we are trying to reach a pH of 4.50, we can find the [H+] concentration by taking the negative log of the pH value:

pH = -log[H+]

4.50 = -log[H+]

[H+] = 3.16 x 10^(-5) M

Now we can use the Henderson-Hasselbalch equation to find the ratio of [HCOO-]/[HCOOH] needed to achieve this pH:

pH = pKa + log([HCOO-]/[HCOOH])

4.50 = 3.75 + log([HCOO-]/[HCOOH])

0.75 = log([HCOO-]/[HCOOH])

Antilog(0.75) = [HCOO-]/[HCOOH]

[HCOO-]/[HCOOH] = 5.62

This means that for every 5.62 molecules of formic acid, we need 1 molecule of sodium formate, which is formed by the reaction of sodium hydroxide and formic acid.

So, we can set up a mole ratio equation to find the amount of NaOH needed:

0.100 M HCOOH x 0.250 L = 0.025 moles HCOOH

0.025 moles HCOOH x (1 mol NaHCOO/5.62 mol HCOOH) = 0.00445 moles NaHCOO

Now we can use the concentration of NaOH to find the volume needed:

0.0500 M NaOH = 0.0500 moles NaOH/L

0.00445 moles NaOH / 0.0500 moles NaOH/L = 0.089 L = 89.0 mL

Therefore, the best choice is B) 420 mL.

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