Question

pH of a solution containing 0.3 M HX and 0.1M X–
(Kb for X– = 1.0 × 10⁻⁵) is –
(a) 5 + log 3(b) 5 – log 3 (c) 9 – log 3(d) 9 + log 3

Answer 1

$\huge\bold\red{Answer:-}$

(b) 5 – log 3

Answer 2

Answer:

5 - log 3 is the pH of the solution.

Explanation:

• This question will be solved by Henderson Hasselbach equation$pH = pKa + log10 ([A]/[HA]$

• pKa is equal to -logKb ⇒ - log(1 × 10⁻⁵) ⇒ 5

• log₁₀ [0.1]/[0.3] ⇒ log₁₀ [1/3] ⇒ -log₁₀(3)

• Assembling values in the equation we get

• pH = 5 - log 3

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