## What is the kw of pure water at 50.0°c, if the ph is 6.630? 5.50 x 10-14 2.13 x 10-14 1.00 x 10-14 2.34 x 10-7 there is not enough informati

Question

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

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The [tex]K_w[/tex] value of water at 50.0u00b0 C is [tex]5.50times 10^{-14}[/tex].

Explanation:

The pH of water at 50.0u00b0 C = 6.630

[tex]pH=-log[H^+][/tex]

[tex]6.630=-log[H^+][/tex]

[tex][H^+]=2.344times 10^{-7}M[/tex]

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

[tex][OH^-]=[H^+]=2.344times 10^{-7}M[/tex]

[tex]H_2Orightleftharpoons H^++OH^-[/tex]

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

[tex]K_w=[H^+][OH^-][/tex]

Substituting the values:

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

The [tex]K_w[/tex] value of water at 50.0u00b0 C is [tex]5.50times 10^{-14}[/tex].

2. Answer is: Kw of pure water at 50.0u00b0C is5.50 x 10-14.
pH = 6.630.
pH = -log
[Hu207a].
[Hu207a] = 10u2227(-pH).
[Hu207a] = 10u2227(-6.63) = 2.34u00b710u207bu2077 M.
[Hu207a]u00b7[OHu207b] = x.
Kw = ?.
nKw =[H
u207a]u00b7[OHu207b].
nKw = xu00b2.
Kw = (
2.34u00b710u207bu2077 M)u00b2.
Kw = 5.50u00b710u207bu00b9u2074 Mu00b2.
Kw is ionic product of water.