An unknown monoprotic acid has a pH of 2.87. Is this acid weak or strong if 22.32 mL of 0.112 M NaOH is required to titrate 25.00 mL of the acid? If the acid is weak, calculate its Ka.

Answer is Ka = 1.8 x 10^-5, but I don't know how to get there. Any help would be appreciated. Thanks!

the initial acid molarity is

... [HA] = .112 * (22.32 / 25.00) = 0.100

[H+] = 10^(-2.87) = .00135 ... this is also [A-]

Ka = [H+][A-] / [HA] = .00135^2 / (.100 - .00135)

remember ... the initial molarity changes because of the dissociation

To determine whether the acid is weak or strong, we need to calculate its Ka value.

First, let's set up the balanced chemical equation for the acid-base reaction between the unknown acid and NaOH:

Unknown acid (HA) + NaOH --> NaA + H2O

Based on the reaction stoichiometry, we can determine the ratio of moles of acid (HA) and moles of NaOH:

moles of HA = moles of NaOH

To find the moles of NaOH, we need to use the concentration and volume of NaOH:

moles of NaOH = concentration of NaOH × volume (in liters)

Given that the volume of NaOH is 22.32 mL and the concentration is 0.112 M, we can calculate the moles of NaOH:

moles of NaOH = 0.112 M × (22.32 mL / 1000 mL/L) = 0.0025 mol

Since the reaction stoichiometry tells us that the moles of acid are equal to the moles of NaOH, the moles of the unknown acid in the reaction are also 0.0025 mol.

Now, we can calculate the molarity (M) of the acid:

Molarity of acid (HA) = moles of acid / volume of acid (in liters)

Given that the volume of the acid is 25.00 mL (or 0.025 L), we can calculate the molarity of the acid:

Molarity of acid (HA) = 0.0025 mol / 0.025 L = 0.1 M

Now, let's use the equation for calculating pH from the concentration of H+ ions:

pH = -log[H+]

Given that the pH of the acid is 2.87, we can calculate the concentration of H+ ions:

[H+] = 10^(-pH)

[H+] = 10^(-2.87) = 1.79 × 10^(-3) M

Since the acid is monoprotic (one acidic hydrogen), the concentration of the acid (HA) and the concentration of H+ ions are equal:

Molarity of acid (HA) = [H+]

Therefore, the molarity of the acid (HA) is also 1.79 × 10^(-3) M.

Now, let's calculate the initial concentration of the undissociated acid (HAi) and the concentration of the dissociated acid (A-) at equilibrium:

HAi = Molarity of acid (HA) = 1.79 × 10^(-3) M
A- = [H+] = 1.79 × 10^(-3) M

To calculate the Ka value, we will use the equation for Ka:

Ka = [A-][H+] / [HAi]

Substituting the known values:

Ka = (1.79 × 10^(-3) M) × (1.79 × 10^(-3) M) / (1.79 × 10^(-3) M) = 1.8 × 10^(-5)

Therefore, the Ka value for the unknown monoprotic acid is 1.8 × 10^(-5).