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

Answer:

The $K_w$ value of water at 50.0° C is $5.50\times 10^{-14}$.

Explanation:

The pH of water at 50.0° C = 6.630

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

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

$[H^+]=2.344\times 10^{-7}M$

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

$[OH^-]=[H^+]=2.344\times 10^{-7}M$

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

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

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

Substituting the values:

$K_w=[2.344\times 10^{-7}M][2.344\times 10^{-7}M]=5.49\times 10^{-14}\approx 5.50\times 10^{-14}$

The $K_w$ value of water at 50.0° C is $5.50\times 10^{-14}$.

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.