How many gm of solid NaOH must be added to 100 ml of a buffer solution which is 0.1 M each w.r.t. Acid HA and salt Na + A − to make the pH of solution 5.5. Given pk a ( HA ) = 5 . (Use antilog (0.5)

Question

How many gm of solid NaOH must be added to 100 ml of a buffer solution which is 0.1 M each w.r.t. Acid HA and salt Na + A − to make the pH of solution 5.5. Given pk a ​ ( HA ) = 5 . (Use antilog (0.5) = 3.16)

TestHub · Accepted Answer

Buffer: 0.1 M HA and 0.1 M A⁻

Target pH = 5.5, pKₐ = 5

Henderson-Hasselbalch:

$p H = p K_{a} + lo g \frac{[ A ^{-} ]}{[ H A ]}$

$5.5 = 5 + lo g \frac{[ A ^{-} ] + x}{[ H A ] - x} ⟹ 0.5 = lo g \frac{0.1 + x}{0.1 - x}$

$\frac{0.1 + x}{0.1 - x} = 3.16 (an t i l o g 0.5)$

$0.1 + x = 3.16 (0.1 - x) ⟹ 0.1 + x = 0.316 - 3.16 x$

$x + 3.16 x = 0.316 - 0.1 ⟹ 4.16 x = 0.216 ⟹ x = 0.0519 M$

Volume = 0.1 L, so moles = $0.0519 \times 0.1 = 0.00519 m o lN a O H$

Mass NaOH = $0.00519 \times 40 \approx 0.208 g \approx 2.08 \times 1 0^{- 1} g$

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