Question

You need to prepare an acetate buffer of pH 6.43 from a 0.723 M acetic acid solution and a 2.47 M KOH solution. If you have 725 mL of the acetic acid solution, how many milliliters of the KOH solution do you need to add to make a buffer of pH 6.43? The pKa of acetic acid is 4.76.

Answer 1

Answer

207.8 ml

Explanation

Henderson -Hasselbalch equation is as follows

pH = pKa + log([A^-]/[HA])

A^- = CH3COO^-

HA = CH3COOH

6.43 = 4.76 + log([CH3COO^-]/[CH3COOH])

log([CH3COO^-]/[CH3COOH]) = 1.67

[CH3COO^-]/[CH3COOH] = 46.77

moles of CH3COO^- / moles of CH3COOH= 46.77

moles of CH3COO^- = 46.77× moles of CH3COOH

moles of buffer = (0.723mol/1000ml) × 725ml = 0.5242mol

moles of CH3COOH + (46.77× moles of CH3COOH) = 0.5242mol

47.77moles of CH3COOH= 0.5242mol

moles of CH3COOH= 0.01097mol

moles of CH3COO^- = 0.5242mol - 0.01097mol = 0.5132mol

Reaction between KOH and CH3COOH

CH3COOH + OH^- -------> CH3COO^- + H2O stoichiometrically, 1mole of CH3COO^- is obtained from 1mole of KOH

So,

To get 0.5132moles of CH3COO^- , 0.5132moles of KOH should be added

Volume of 2.47M KOH solution containing 0.5132moles of KOH = (1000ml /2.47mol)× 0.5132mol = 207.8ml

Therefore

Volume of 2.47M KOH needed = 207.8