Question

A buffer solution contains 0.387 M NaH2PO4 and 0.394 M Na HPO4. Determine the pH change when 0.094 mol NaOH is added to 1.00 L of the buffer. pH after addition - pH before addition = pH change

Answer 1

1) Before addition of NaOH

pH of buffer = pka + log(Na2HPO4/NaH2PO4 )

pka2 of H3PO4 = 7.2

[Na2HPO4] = 0.394 M

[NaH2PO4 ] = 0.387 M

pH = 7.2 + log(0.394/0.387)

pH = 7.2

after addition of NaOH

pH of buffer = pka + log(Na2HPO4 + NaOH/NaH2PO4 - NaOH)

pka2 of H3PO4 = 7.2

[Na2HPO4] = 0.394 M

[NaH2PO4 ] = 0.387 M

[NaOH] added = 0.094 M

pH = 7.2 + log((0.394+0.094)/(0.387-0.094))

pH = 7.42

pH change = 7.42-7.2 = 0.22

2)
Before addition of NaOH

pH of acidic buffer = pka + log(ClO-/HClO)

pH = ?

pka of HClO = 7.54

[ClO-] = 0.235 M

[HClO] = 0.498 M

pH = 7.54+log(0.235/0.498)

pH = 7.21

after addition of NaOH

pH of acidic buffer = pka + log((ClO- + NaOH)/(HClO-NaOH))

pH = ?

pka of HClO = 7.54

[ClO-] = 0.235 M

[HClO] = 0.498 M

[NaOH] = 0.117 M

pH = 7.54+log((0.235+0.117)/(0.498-0.117))

pH = 7.5

Change in pH = 7.5 - 7.21 = 0.29