You visited us 1 times! Enjoying our articles? Unlock Full Access!

pH of a Solution

Kw of pure wa...

Question

$K_{w}$ of pure water is $10^{- 12}$ at $60^{o} C$ . The $p H$ of pure water at $60^{o} C$ is:

Open in App

Solution

$K_{w} = [H^{+}] [O H^{-}] = 10^{- 12}$
$[H^{+}] = [O H^{-}] = 10^{- 6}$
$p H = - l o g [H^{+}] = - l o g 10^{- 6} = 6$ .

Suggest Corrections

Similar questions

60^{o} C

[H_{3} O^{+}] = 10^{- 6.7} m o l / l i t

K_{W}

60^{o} C

K_{w}

H_{2} O

9 \times 10^{- 14}

60^{o} C

p H

60^{o} C

(l o g 3 = 0.47)

K_{w}

H_{2} O

9.62 \times 10^{- 14}

60^{o} C

p H

60^{o} C

K_{w}

9.62 \times 10^{- 14}

60^{o} C

p H

K_{w}

2 H_{2} O ⇌ H_{3} O^{+} + O H^{-}

10^{- 14}

25^{o} C

9.62 \times 10^{- 14}

60^{o} C

p H

60^{o} C

\sqrt{9.60} = 3.10, log 3.10 = 0.49

Join BYJU'S Learning Program