To a 25.00-mL volumetric flask, a lab technician adds a 0.400 g sample of a weak

Question

To a 25.00-mL volumetric flask, a lab technician adds a 0.400 g sample of a weak monoprotic acid, HA, and dilutes to the mark with distilled water. The technician then titrates this weak acid solution with 0.0955 M KOH. She reaches the endpoint after adding 42.79 mL of the KOH solution. Determine the number of moles of the weak acid in the solution. Determine the molar mass of the weak acid. After the technician adds 18.15 mL of the KOH solution, the pH of the mixture is 4.17. Determine the pK_a of the weak acid. pK_a =

Explanation / Answer

moles of acid = moles of base

moles of base = 0.0955 x 42.79 / 1000 = 4.086 x 10^-3

moles of acid = 4.086 x 10^-3

moles = mass / molar mass

4.086 x 10^-3 = 0.400 / molar mass

molar mass = 97.9 g / mol

moles of base = 0.0955 x 18.15 / 1000 = 1.73 x 10^-3

HA   + KOH ------------------------> KA + H2O

4.086    1.73                                 0         0

2.53        0                                    1.73

pH = pKa + log [KA / HA]

4.17 = pKa   + log (1.73 / 2.53)

pKa = 4.34

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