PH of the solution produced by mixing equal volumes of 2 - 10-3 M HClO 4 and 1 - 10-2 M KClO 4 will be-

Let volume of each solution be V ml Moles of [H^+] = 2 × 10^ 3× V moles Final volume, V_f = 2V ml ∴ [H^+] = 2 × 10^ 3× V2V = 10^ 3 M We know, pH = log[H^+] ⇒ p ...

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Standard XII

Chemistry

Henderson Hassleback Equation

pH of the sol...

Question

$p H$ of the solution produced by mixing equal volumes of $2 \times 10^{- 3} M H C l O_{4}$ and $1 \times 10^{- 2} M K C l O_{4}$ will be:

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Solution

Let volume of each solution be $V ml$
Moles of $[H^{+}] = 2 \times 10^{- 3} \times V moles$
$Final volume, V_{f} = 2 V ml$
$∴ [H^{+}] = \frac{2 \times 10^{- 3} \times V}{2 V} = 10^{- 3} M$
We know,
$p H = - log [H^{+}]$
$\Rightarrow p H = - l o g 10^{- 3} = 3$
Since $K C l O_{4}$ is a neutral salt, thus it does not affect the $p H$ of the solution.

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pH of the solution produced by mixing equal volumes of 2×10−3 M HClO4 and 1×10−2 M KClO4 will be:

Let volume of each solution be V ml Moles of [H+]=2×10−3×V moles Final volume, Vf=2V ml ∴[H+]=2×10−3×V2V=10−3 M We know, pH=−log[H+] ⇒pH=−log10−3=3 Since KClO4 is a neutral salt, thus it does not affect the pH of the solution.

$p H$ of the solution produced by mixing equal volumes of $2 \times 10^{- 3} M H C l O_{4}$ and $1 \times 10^{- 2} M K C l O_{4}$ will be: