Question

To prepare a buffer of pH 8.26 amount of (NH4)2 SO4 to be added to 500 mL of 0.01 M NH4OH solution [pKa (NH4+) = 9.26] is

• A. 0.05 mole
• B. 0.025 mole
• C. 0.10 mole
• D. 0.005 mole

Answer 1

Answer: (B)

0.025 mole

Answer 2

For the buffer solution of NH3 & NH4+

pH = pKa + log $\Rightarrow$ 8.26 = 9.26 + log $\frac { ( 500 \times 0.01 ) } { \text { m.moles of } \mathrm { NH } _ { 4 } ^ { + } }$

$\therefore$ moles of (NH4)2 SO4 required = 0.025.