Question

kw of water at 373K is 1*10-12 what will be the pH of water at 373K? is water acidic , basic or neutral at this temp?

Answer 1

 

T=373K

Kw=1. 10-12

PH=?

Relation lies between [H+] + and [OH-]:

[H+] + [OH-] = 10 -14

[H+]=10 –PH and

PH + POH = 14

PH = 14 – POH

Than PH = 14 + log [OH-]

10 14+log [OH-] * [OH-] = 10-14

14 + log [OH-] 2 = 10-14

Logx14+log [OH-] 2=logx10power of -14

14+ [OH-] 2 =10-14

[OH-] 2= 14

[OH-]= √14

[H+] [OH-]=10-14 → [OH-] = 3, 74

[H+] 3, 74= 10-14

[H+]=10-14/ 3, 74

[H+]=2, 67.10-15

PH= - log [H+]

PH= -log 2, 67 10-15

PH= 14

Water is basic   because PH=14                

                                                                               Ruben Maseka

9c170567a48afd74a43eccb5f332b67d.docx

Answer 2

Answer : The pH of water at 373 K is, 6 and the water is neutral at this temperature.

Explanation : Given,

$K_w=1\times 10^{-12}$

First we have to calculate the concentration of $H^+$

The expression for dissociation constant of water will be,

$K_w=[H^+][OH^-]$     ........(1)

Let us assume that the concentration of $H^+$ is, 'x'

As we know that,

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

Now put all the given values in equation (1), we get

$1\times 10^{-12}=(x)\times (x)$

$1\times 10^{-12}=x^2$

$x=1\times 10^{-6}$

The concentration of $H^+$ = $1\times 10^{-6}$

Now we have to calculate the pH of water.

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

$pH=-\log (1\times 10^{-6})$

$pH=6$

Therefore, the pH of water at 373 K is, 6 and the water is neutral at this temperature.