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pH the Power of H

At 25 0 C d...

Question

At $25^{0} C$ dissociation constants of acid HA and base BOH in aqueous solution are same. the pH of 0.01 M solution of HA is 5. The pOH of $10^{- 4}$ M solution of BOH at the same temperature is:

3.5

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Solution

The correct option is D 6
$p^{H} = 5$

$\Rightarrow [H^{+}] = 10^{- 5}$

$\Rightarrow \sqrt{c \times K_{a}} = 10^{- 5}$

$\Rightarrow 10^{- 2} \times K_{a} = 10^{- 10}$

$\Rightarrow K_{a} = 10^{- 8}$

$K_{a}$ of $B O H = K_{a}$ of $H A = 10^{- 8}$

$∴ α$ for $B O H = \sqrt{\frac{K_{a}}{C^{1}}}$

$= \sqrt{\frac{10^{- 8}}{10^{- 4}}}$

$= 0.01 α 0.1$

Now $[O H^{-}] = c α$

$= 10^{- 4} \times 10^{- 2}$

$= 10^{- 6}$

$∴ p O H = 6$

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Similar questions

Q. At

25^{\circ} C

disociation constant of acid

H A

and base

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in aqueous solution are same. The

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Q. At

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the ionization constants of

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the dissociation constant

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At 250C dissociation constants of acid HA and base BOH in aqueous solution are same. the pH of 0.01 M solution of HA is 5. The pOH of 10−4 M solution of BOH at the same temperature is:

The correct option is D 6pH=5 ⇒[H+]=10−5 ⇒√c×Ka=10−5 ⇒10−2×Ka=10−10 ⇒Ka=10−8 Ka of BOH=Ka of HA=10−8 ∴α for BOH=√KaC1 =√10−810−4 =0.01α0.1 Now [OH−]=cα =10−4×10−2 =10−6 ∴pOH=6

At $25^{0} C$ dissociation constants of acid HA and base BOH in aqueous solution are same. the pH of 0.01 M solution of HA is 5. The pOH of $10^{- 4}$ M solution of BOH at the same temperature is: