Question

A buffer is prepared by mixing 206 mL of 0.452 M HCl and 0.470 L of 0.400 M sodium acetate. What is the pH? How many grams of KOH must be added to 0.500 L of the buffer to change the pH by 0.16 units?

Answer 1

moles HCl = 0.206 L x 0.452 M= 0.09311 moles
moles sodium acetate = 0.470 L x 0.400 M = 0.188 moles

CH3COO- + H+ => CH3COOH
moles CH3COO- = 0.188 - 0.09311= 0.09489 moles
moles CH3COOH = 0.09311
total volume = 0.676 L
[CH3COO-]= 0.09489/ 0.676 =0.14 M
[CH3COOH]= 0.09311/ 0.676 = 0.1377 M
pKa = 4.74
pH = 4.74+ log (0.140/0.1377)= 4.74

And how much KOH must be added to 0.5 L of the solution to change the pH by 0.16 units

pH = 4.74 + 0.16 = 4.9

4.9 = 4.74 + log ([C2H3O2-] / [HC2H3O2])
log ([C2H3O2-] / [HC2H3O2]) = 4.9 – 4.74 = 0.16

KOH will decrease the affect of the HCl. Let’s go back and determine the moles of HCl that produce pH = 4.9
Let x = moles of HCl
x moles of HCl will react with x moles of NaC2H3O2 to produce x moles of HC2H3O2 and x moles of NaCl.
[HC2H3O2] = x ÷ 0.676
The excess NaC2H3O2 = 0.188 – x
Moles of C2H3O2- = 0.188 – x
[C2H3O2-] = (0.188 – x) ÷ 0.676

log ([C2H3O2-] / [HC2H3O2]) = log [(0.188 – x) / x]
log [(0.188 – x) / x] = 0.16
x = 0.0765
This the moles of HCl to produce pH = 4.9
Original moles of HCl = 0.09311
Reduction of moles of HCl = 0.09311 – 0.0765 = 0.01661
1 mole of KOH will react with 1 mole of HCl.
We need to add 0.01661 moles of KOH in the 0.676 liters of solution.

How much KOH must be added to .5L of solution?
0.0164705 * 0.5/0.704 = 0.0117 mole of KOH