Calculate the pOH of a solution at 25- that contains 1- 10-10 M of hydronium ions- i-e- H3O-

The correct option is A 4Given [H_3O^+ #160; ]=1× 10^ 10 M At 25^∘ C, #160; [H_3O^+ #160; ] [OH^ #160; ] #160;=10^ 14 ∴ [OH^ #16 ...

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Standard VII

Chemistry

Nersnt Equation

Calculate the...

Question

Calculate the pOH of a solution at $25^{\circ}$ that contains $1 \times 10^{- 10} M$ of hydronium ions, i.e. $H_{3} O^{+}$ .

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Solution

The correct option is A 4
Given $[H_{3} O^{+}] = 1 \times 10^{- 10} M$
At $25^{\circ} C, [H_{3} O^{+}] [O H^{-}] = 10^{- 14}$
$∴ [O H^{-}] = \frac{10^{- 14}}{10^{- 10}} = 10^{- 4}$
Now, $[O H^{-}] = {10^{- p}}^{O H} = 10^{- 4} = 10^{- p O H}$
$∴ p O H = 4$

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Calculate the pOH of a solution at 25∘ that contains 1×10−10M of hydronium ions, i.e. H3O+.

The correct option is A 4Given [H3O+]=1×10−10MAt 25∘C,[H3O+][OH−]=10−14∴[OH−]=10−1410−10=10−4Now, [OH−]=10−pOH=10−4=10−pOH∴pOH=4

Calculate the pOH of a solution at $25^{\circ}$ that contains $1 \times 10^{- 10} M$ of hydronium ions, i.e. $H_{3} O^{+}$ .

The correct option is A 4
Given $[H_{3} O^{+}] = 1 \times 10^{- 10} M$
At $25^{\circ} C, [H_{3} O^{+}] [O H^{-}] = 10^{- 14}$
$∴ [O H^{-}] = \frac{10^{- 14}}{10^{- 10}} = 10^{- 4}$
Now, $[O H^{-}] = {10^{- p}}^{O H} = 10^{- 4} = 10^{- p O H}$
$∴ p O H = 4$