Question

The value of the ion product constant for water, (Kw) at 60°C is 9.6 x 10-14 M². What is the
[H30+] of a neutral aqueous solution at 60°C and an aqueous solution with a pH = 7.0 at 60°C
are respectively?
(a) 3.1x 10^-8, acidic
(C) 3.1x10^-8, basic
(b) 3.1x 10^-7, neutral
(d) 3.1x 10^-7, basic​

Answer 1

Answer:

Solution:- (D) 3.1×10−7, basic

Given:-

Kw=9.6×10−14M2

As we know that,

for a neutral solution,

[H3O+]=[OH−]

Now,

Kw=[H3O+]×[OH−]

⇒Kw=[H3O+]2(∵[H3O+]=[OH−])

⇒[H3O+]2=9.6×10−14

_______

⇒[H3O+]= /9.6×10−14≈3.1×10−7

Hence at 60℃, for a neutral aqueous solution, the [H3O+] is 3.1×10−7.

Now,

pH of the neutral solution-

pH=−log[H3O+]

pH=−log(3.1×107)

⇒pH=(7−log(3.1))=6.51

Since pH of a neutral solution is less than 7, the aqueous solution with a pH=7 is basic in nature

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