Question

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 K_{a} for the acid?

Answer #1

**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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way, you would help me a lot, thank you.**

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