Question

# A 1.26 gram sample of an unknown monoprotic acid is dissolved in 50.0 mL of water...

A 1.26 gram sample of an unknown monoprotic acid is dissolved in 50.0 mL of water and titrated with a a 0.306 M aqueous potassium hydroxide solution. It is observed that after 12.7 milliliters of potassium hydroxide have been added, the pH is 3.193 and that an additional 19.3 mL of the potassium hydroxide solution is required to reach the equivalence point.

(1) What is the molecular weight of the acid? ____ g/mol

(2) What is the value of Ka for the acid?

The moles of KOH required to reach the equivalence point are calculated:

n KOH = M * V = 0.306 M * 0.0193 L = 0.006 mol

The moles of KOH are equal to the moles of acid, the molar weight is calculated:

MM = g / n = 1.26 g / 0.006 mol = 210 g / mol

The moles of KOH added to the 12.7 mL are calculated:

n KOH = 0.306 * 0.0127 = 0.004 moles

The remaining moles of acid are calculated:

n Acid = 0.006 - 0.004 = 0.002 mol

The pKa is calculated by the equation of henderson hasselbach cleared:

pKa = pH - log (n Salt / n Acid) = 3.193 - log (0.004 / 0.002) = 2.89

The Ka is calculated:

Ka = 10 ^ -pKa = 10 ^ 2.89 = 1.29x10 ^ -3

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