A sample of 100 mL of 0.10 M acid HA ( K _a = 1.0×10^{- 5}) is titrated with standard 0.100 M KOH. How many mL of KOH will have been added when the pH in the titration flask is 5.00?

Question

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To solve this problem, we need to use the balanced chemical equation for the reaction between acid HA and base KOH:

HA + KOH → K+A + H2O

In this reaction, one mole of acid reacts with one mole of base. The balanced equation tells us that the acid reacts with the base in a 1:1 ratio, so the moles of acid will be equal to the moles of base at the equivalence point.

At the equivalence point, all of the acid will have reacted with the base, leaving only the conjugate base A- and hydroxide ions OH- in the solution. The pH of the solution will be determined by the concentration of OH-, which can be calculated from the concentration of the added base and the volume of the solution.

We can use the formula for the equilibrium constant of the reaction to find the concentration of OH- at the equivalence point:

Kw = [H3O+][OH-] = 1.0 × 10^-14

At the equivalence point, [HA] = 0 and [A-] = [KOH]. Let x be the volume of KOH required to reach the equivalence point.

The initial concentration of HA is 0.10 M, and the initial volume is 100 mL = 0.1 L. At the equivalence point, the volume of the solution will be 100 mL + x mL.

From the balanced chemical equation, we know that the moles of acid and base are equal at the equivalence point:

nHA = nKOH

MHA × VHA = MKOH × VKOH

0.10 M × 0.1 L = 0.100 M × x L

x = 1.0 L

So, the total volume of the solution at the equivalence point is 1.1 L.

The concentration of OH- at the equivalence point can be found using the formula:

[OH-] = MKOH × VKOH / Vtot

[OH-] = 0.100 M × 1.0 L / 1.1 L

[OH-] = 0.09091 M

The pH of the solution at the equivalence point can be calculated using the formula:

pH = 14 - pOH

pH = 14 - log[OH-]

pH = 14 - log(0.09091)

pH = 5.02

So, the closest choice is D) 50 mL, which is the volume of added base at the midpoint between the initial pH of the solution and the pH at the equivalence point. At the midpoint of the titration, half of the acid has been neutralized, so the concentration of HA is equal to the concentration of A-, and the pH is equal to the pKa of the acid. For a weak acid like HA, the pKa is close to the initial pH of the solution, which is 2.00. The pH at the midpoint is:

pH = pKa + log([A-] / [HA])

2.00 = -log(Ka) + log([KOH] / [HA])

-log(Ka) = 2.00 - log(0.1 / Ka)

-log(Ka) = log(10Ka) - 2.00

Ka = 10^-5

[HA] = [A-] = 0.05 M (half of the initial concentration)

pH = pKa + log([A-] / [HA])

pH = -log(10^-5) + log(0.05 / 0.05)

pH = 5.00

So, when the pH in the titration flask is 5.00, we are at the midpoint of the titration, and 50 mL of KOH have been added.

A Sample of 100 ML of 0