Question

pH of $10^{-8}$M HCl solution is

Hint :- Answer should be "6.99".

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Answer 1

Answer:-

pH of 10⁻⁸ M HCl solution is 6.99 .

Explanation:-

H₂O ⇌ H⁺ + OH⁻

Since HCl is a strong acid, it ionizes completely.

Now, we know that :-

H⁺[HCl] = 1 × 10⁻⁸

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Suppose [OH⁻] = x

∴ H⁺ = x + 1 × 10⁻⁸

Kw = [H⁺][OH⁻] = 1 × 10⁻¹⁴

⇒ (x + 1 × 10⁻⁸)(x) = 1 × 10⁻¹⁴

⇒ x² + 1 × 10⁻⁸x = 1 × 10⁻¹⁴

⇒ x² + 1 × 10⁻⁸x - 1 × 10⁻¹⁴ = 0

⇒ x = 9.5 × 10⁻⁸

[H⁺] = x + 1 × 10⁻⁸

= 9.5 × 10⁻⁸ + 1 × 10⁻⁸

= 1.05 × 10⁻⁷ mol L⁻¹

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pH = - log[H⁺]

⇒ pH = - log[1.05 × 10⁻⁷]

⇒ pH = 7 - log1.05

⇒ pH = 6.98 ≈ 6.99

Answer 2

Question :

⠀⠀⠀⠀▪︎ pH of $\sf 10^{-8}$M HCl solution is _________ .

⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀

⠀⠀⠀⠀⠀⠀⠀❒ Finding the pH of $10^{-8}$M HCl :

⠀⠀⠀⠀⠀⠀⠀There are two different source of HCl in aqueous solution :

• Acid $\: \sf H^+$ = $\sf 10^{-8}$
• Water $\: \sf H^+$ = $\sf 10^{-7}$ [ pH of Water is 7 ]

⠀⠀⠀⠀⠀━━━ Total Hydrogen ions $\: \sf [ \ H^+ \ ]$ in aqueous solution is :

$\qquad :\implies \sf \underbrace{10^{-8}}_{HCl } + \underbrace{10^{-7}}_{H_2O }$

$\qquad :\implies \sf 10^{-8} + 10^{-7}$

$\qquad :\implies \sf 10^{-7} (10^{-1} + 1 )$

$\qquad :\implies \sf 10^{-7} (\dfrac{11}{10} )$

$\qquad :\implies \sf 10^{-7} (1.1)$

$\qquad :\implies \sf 1.1 \times 10^{-7}$

$\qquad :\implies \bf 1.1 \times 10^{-7} M$

$\dag\:\:\sf{ As,\:We\:know\:that\::}$

$\qquad \dag\:\:\bigg\lgroup \sf{pH \:: - log [ H^+ ] }\bigg\rgroup$

Here ,

• $\sf H^+$ = $\qquad \bf 1.1 \times 10^{-7} M$

⠀⠀⠀⠀⠀⠀$\underline {\boldsymbol{\star\:Now \: By \: Substituting \: the \: known \: Values \::}}$

$\qquad :\implies \sf pH \:= - log [ H^+ ]$

$\qquad :\implies \sf pH \:= - log [ 1.1 \times 10^{-7} ]$

$\qquad :\implies \sf pH \:= 7 - log 1.1$

$\qquad :\implies \sf pH \:= 7 - log 1.1$

$\qquad :\implies \bf pH \:= 6.99$

$\qquad \longmapsto \frak{\underline{\purple{\:pH =6.99 }} }\:\:\bigstar$

Therefore,

⠀⠀⠀⠀⠀$\therefore {\underline{ \mathrm {\:pH \:of\: 10^{-8}M\: HCl \:solution \;is\:\bf{6.99}}}}$

⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀