hello
how would i do this question 14c) and 14d) i know 14 a and b are increase
for 14c) if both h30+ and OH- ions concerntration increases how does that equate to an overall ph decrease? cause i know the increase h30+ concerntration would account for the decreased ph, but wouldnt the increase oh- then counteract that?
for 14d) if 14c) is in fact ph decrease, wouldnt that mean the water becomes acidic, but then i know that h30+ concerntration = oh- concerntration therefore it has to be neutral, but then the ph decreased?
so confused?
A solution is considered as neutral when
pH=7 at 25C . When you increase, or decrease the temperature outside 25C, a solution is considered neutral when [H+] = [OH-]. Hence, you can have a solution of pH = 6 at a given temperature but still called neutral.
Kw = [H30+] x [OH-] = 10^(-14) M
at 25CWhen you increase the temperature, forward reaction is favoured, hence, you obtained more of both [H30+] and [OH-]. This will change the value of Kw (that's why like all equilibrium constant, Kw is temperature dependent).
As mentioned above, because in a neutral solution, you have equal amount of hydronium and hydroxide, the Kw
at any given temperature is
Kw= [H30+] x [OH-] = [H30+]^2
Since Kw is increasing when you apply heat, let's make up a larger Kw value and see how that changes the pH value.
New Kw = 51.3 x 10^(-14) M
[H3O+] = *sqrt* 51.3 x 10^(-14)
[H30+] = 7.16 x 10 ^ (-7)
Hence, pH = 6.14
In brief, even if you increase the temperature, the solution is still neutral despite the fact that it's pH is decreasing. pH 7 of a neutral solution only holds at 25C.
Correct me if I'm wrong.
[Beaten]