Question

What is the kw of pure water at 50.0°c, if the ph is 6.630? 5.50 × 10-14 2.13 × 10-14 1.00 × 10-14 2.34 × 10-7 there is not enough information to calculate the kw?

Answer 1

The
`K_w` value of water at 50.0° C is
`5.50* 10^(-14)`.

Step-by-step explanation:

The pH of water at 50.0° C = 6.630

`pH=-\log[H^+]`

`6.630=-\log[H^+]`

`[H^+]=2.344* 10^(-7)M`

Since, water is a neutral compound wuith equal number of hydrogen ions and hydroxide ions.

`[OH^-]=[H^+]=2.344* 10^(-7)M`

`H_2O\rightleftharpoons H^++OH^-`

The expression for an ionic product of water is given as:

`K_w=[H^+][OH^-]`

Substituting the values:

`K_w=[2.344* 10^(-7)M][2.344* 10^(-7)M]=5.49* 10^(-14)\approx 5.50* 10^(-14)`

The
`K_w` value of water at 50.0° C is
`5.50* 10^(-14)`.

Answer 2

Answer is: Kw of pure water at 50.0°C is 5.50 × 10-14.
pH = 6.630.
pH = -log[H⁺].
[H⁺] = 10∧(-pH).
[H⁺] = 10∧(-6.63) = 2.34·10⁻⁷ M.
[H⁺] · [OH⁻] = x.
Kw = ?.
Kw = [H⁺] · [OH⁻].
Kw = x².
Kw = (2.34·10⁻⁷ M)².
Kw = 5.50·10⁻¹⁴ M².
Kw is ionic product of water.