A sample of 0.2505 g of an unknown monoprotic acid was dissolved in 25.0 mL of water and titrated with 0.0950 M . The acid required 30.0 mL of base to reach the equivalence point.

After 15.0 mL of base had been added in the titration, the pH was found to be 6.50. What is the K_{a} for the unknown acid?

We’re being asked to **calculate the K _{a} of the given monoprotic acid**.

Recall that at the ** equivalence point** of a titration:

$\overline{){\mathbf{moles}}{\mathbf{}}{\mathbf{acid}}{\mathbf{=}}{\mathbf{moles}}{\mathbf{}}{\mathbf{base}}}$

Also, recall that **moles = molarity × volume**.

This means:

$\overline{){\left(\mathbf{MV}\right)}_{{\mathbf{acid}}}{\mathbf{=}}{\left(\mathbf{MV}\right)}_{{\mathbf{base}}}}$

**The monoprotic acid will have 1 equivalence point:**

**HA → H ^{+} + A^{-} **

** (MA) _{acid} = (MV)**

We are given:

[HA]=**0.0950 M**

Volume of HA = **25 mL**

Volume of A^{- }to reach equivalence point=**30 mL**

Recall: A^{-}

**1 mL = 10 ^{-3} L**