Question

Sample Buffering Capacity Problem Suppose you have 600 ml of a sodium phosphate buffer where Concentration = .25M Ratio of Acid : Base = 2:3 PK, of Sodium dihydrogen phosphate = 7.21 Next we add 150 ml of .2M HCL How much does the pH change after this addition?

Answer 1

The acid here is NaH₂PO₄ and base is Na₂HPO₄. Now, according to the question,

$[Na_2HPO_4]+[NaH_2PO_4] = 0.25M$

$\frac{[NaH_2PO_4]}{[Na_2HPO_4]} = \frac{2}{3}$

$Thus,\frac{[NaH_2PO_4]+[Na_2HPO_4]}{[Na_2HPO_4]} = \frac{2+3}{3}=\frac{5}{3}$

$Thus,\frac{0.25}{[Na_2HPO_4]}=\frac{5}{3}$

$Thus, [Na_2HPO_4]=\frac{0.25\times 3}{5}=0.15$

$And\, [NaH_2PO_4]=(0.25-0.15)=0.10$

From, Henderson-Hasselbalch equation:

$pH=pK_a+log \frac{[Na_2HPO_4]}{[NaH_2PO_4]}=7.21+log\frac{0.15}{0.10}=7.21+0.18=7.39$

Now, mL x M = mL x (mol / L) = mL x (mol / 10³ mL) = 10^-3 mol = mmol

So, 600 mL 0.10 M NaH₂PO₄ = 60 mmol

600 mL 0.15 M Na₂HPO₄ = 90 mmol

acid added = 150 mL 0.2 M HCl = 30 mmol

So, now mmoles of acid is increased to (60+30) mmol = 90 mmol

mmoles of base decreased to (90-30) mmol = 60 mmol

So now

$pH=7.21+log\frac{60}{90}=7.21-.18=7.03$

So, change in pH = (7.39 - 7.03) = 0.36 (in negative direction).