Question

(i)At 60° C , if water has [ H+] = 10^-5 mol/litre,then type of solution is ______

(ii)The pH of a solution is 5.0.Is hydrogen ion concentration is decreased 100 times,then solution will be ____.

solve the above two questions ,write the steps involved and formula used.

Option is Acidic/Basic

Answer 1

pH = -log {concentration of hydrogen ion}
at 90°c pH neutral mark is 6 since kw = 10^-12
thus equilibrium concentration of hydrogen = 10^-6
thus pH equilibrium at 90°c = 6
and at 25° pH = 7
60 lies between 25°c and 90°c
and given concentration is 10^-5
thus given pH = 5
so acidic
in 2nd case
if temperature is 25°c
then
given pH = 5
means concentration of hydrogen ion = 10^-5
decrease 100 times became 10^-7
now new pH = 7
so solution will be neutral
if at 60°c then it will be basic

Answer 2

. At higher temperature the extent of ionization increases and hence the concentration of H+ and OH- increases.

Here concentration of H + = 10-5 M

Hence the pH of this solution will be 5

[H+] = 10-5

pH = -log [H+]

pH = -log 10-5

pH = 5

Hence theoretically it is acidic solution but it will remain neutral as the concentration of H + and OH- in the solution will remain not same . Hence at 60 degree pH of a acidic substance will become 5.


2. A strong acid disolved in water it dissociates completely to cation and anian. The cation always a proton (H+atom)

pH value always related to H+ concentration in a solution.

If you know logarithms when someone gives you pH value you can find H+ concentration.

And he gives you H+ concentration you can find the pH value easily.

[H+] stands for H+ concentration

pH value is given by this equation

pH = -log[H+]

Now we find what is the H+ concentration when the pH value is 5.

5 = -log[H+]

-5 = log[H+]

Antilog(-5)= [H+]

10^(-5) = [H+]

concentration of H+ is 10^(-5)moldm³-

We know if any solution diluted two times,the concentration is reduced by 2 times.

Thus when diluted by 100. times the concentration is reduced by 100 times.

We know the concentration of given acedic solution. It is 10^(-5)moldm³-.

When this solution dilute by 100 times,the H+ concentration reduced by 100 times.The new concentration is 10^(-7)moldm³-

New pH value is

pH = - log[ H+]

pH = -log[ 10^(-7)]

pH = -(-7)

pH = 7

Hence it is neutral.