A 25.0-mL sample of 0.35 M HCOOH is titrated with 0.20 M KOH. What is the pH of the solution after 25.0 mL of KOH has been added to the acid? K_a = 1.77 × 10^-4

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Quizplus · Accepted Answer

Before adding any KOH, the solution is just 0.35 M HCOOH. This is a weak acid, so we need to use the Ka expression to find the initial concentration of H+ ions:

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

At equilibrium, the concentration of HCOO- and H+ will be equal, so we can write:

Ka = [H+]^2 / [HCOOH]

Solving for [H+], we get:

[H+] = sqrt(Ka*[HCOOH])

[H+] = sqrt(1.77x10^-4 * 0.35) = 0.012 M

Now we can use the titration information to find out how much of the HCOOH has been neutralized. When we add 25.0 mL of 0.20 M KOH, we are adding:

0.020 L * 0.20 mol/L = 0.004 mol KOH

Since the stoichiometry of the reaction is 1:1 between KOH and HCOOH, we know that 0.004 mol of HCOOH has been neutralized. To find out how much HCOOH is left, we can subtract:

moles HCOOH at start - moles HCOOH neutralized = moles HCOOH remaining

0.35 mol/L * 0.0250 L - 0.004 mol = 0.00625 mol remaining

Now we need to find the concentration of HCOOH and HCOO- in the solution at this point. We know that the total volume is 25.0 + 25.0 = 50.0 mL = 0.0500 L. So the new concentration of HCOOH is:

[HCOOH] = 0.00625 mol / 0.0500 L = 0.125 M

And the new concentration of HCOO- is also 0.125 M (since KOH is a strong base and completely dissociates). Now we can use the Kb expression for HCOO- to find the concentration of OH- ions:

Kb = [OH-][HCOOH] / [HCOO-]

Kb = [OH-]^2 / [HCOO-]

Solving for [OH-], we get:

[OH-] = sqrt(Kb*[HCOO-])

Kb is the same as Kw/Ka, so we can use:

Kw/Ka = 1.00x10^-14 / 1.77x10^-4 = 5.65x10^-11

Kb = 5.65x10^-11 / 1.77x10^-4 = 3.19x10^-7

[OH-] = sqrt(3.19x10^-7 * 0.125) = 3.00x10^-4 M

Finally, we can use the fact that Kw = [H+][OH-] to find the pH:

Kw = 1.00x10^-14 = [H+][OH-]

[H+] = 1.00x10^-14 / [OH-] = 1.00x10^-14 / 3.00x10^-4 = 3.33x10^-11 M

pH = -log[H+] = -log(3.33x10^-11) = 3.88

A 250-ML Sample of 0