You need to prepare an acetate buffer of pH 6.43 from a 0.723 M acetic acid...
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.
Homework Answers
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
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