Question

At 5.0°C, the value of Kw for the equilibrium shown above is 1.9×10−15 and the value of pKw is 14.73. Based on this information, which of the following is correct for pure water at this temperature?

Answer 1

A

Step-by-step explanation:

Kw value is equal to [H30+][OH-]

In the pKw, p is -log, so we need to reverse the negative log via the following:

Kw=
`10^(-pKw)`

Kw=
`10^(-14.73)`

Kw= 1.9x
`10^(-15)`

Now in pure water, [H3O+]=[OH-]

We can derive the following equation from the given, Kw=
`[H3O+] ^(2)`since both hydronium and hydroxide have equal concentrations.

Substitute:

[H3O+]=
`\sqrt{1.9 x 10^(-15) }`

(we use square root as [H3O+] is squared)