Equilibrium Question 318
At $ 80^{o}C, $ distilled water has $ [H_3{O^{+}}] $ concentration equal to $ 1\times {10^{-7}}mole/litre. $ The value of $ K_{w} $ at this temperature will be [CBSE PMT 1994; RPMT 2000; AFMC 2001; AIIMS 2002; BHU 2002]
Options:
A) $ 1\times {10^{-6}} $
B) $ 1\times {10^{-9}} $
C) $ 1\times {10^{-12}} $
D) $ K _{sp} $
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Answer:
Correct Answer: C
Solution:
$ K_{w}=[H_3{O^{+}}][OH^{-}] $ Concentration of $ H_3{O^{+}} $ in distilled water $ =1\times {10^{-7}} $ mol/l. Now $ [H_3{O^{+}}]=[OH^{-}]$
$ K_{w}=[1\times {10^{-7}}]\times [1\times {10^{-7}}]=1\times {10^{-14}} $ .
BETA
