What is the ph of a solution that contains 1.0 l of 0.10 m ch3cooh and 0.080 m nach3coo after 0.03 moles of naoh added?

Answer: pH = 4.996

Explanation:

No of moles = molarity x volume

:• no of moles of CH3COOH = 0.1M x 0.1L

n (CH3COOH) = 0.1mol

Since 0.03mole of NaOH is added, then 0.03 mole of CH3COOH will be converted to the conjugate.

Therefore, Moles of CH3COOH becomes,

0.1 - 0.03 = 0.07 mol

Subsequently, the moles of CH3COONa increases and becomes,

0.08 + 0.03 = 0.11 mol

Using the Hendersom-Hasselbach equation,

pH = pKa + log [Moles of conjugate: moles of Ch3COOH]

From literature, pKa of Ch3COOH is 4.8

Thus,

pH = 4.8 + log [0.11/0.07]

pH = 4.8 + 0.1963

pH = 4.996