Ph of an aqueous solution of nacl at 85 degree c should be
Answers
It should be less than 7
The time when we add NaCl into water, it will completely dissociate into Na+ and Cl- ions ( salt hydrolysis ).
Parallelly water will build up it's own equilibrium
H2O → OH- + H+ K = 10^(-14 ) at 25℃
Na+ will combine with OH- whereas Cl- will combine with H+ , but as NaOH and HCl are strong base and acid simultaneously ,they will immediately dissociates into its ionic form completely
So in medium, we will have equal concentration of H+ = OH-
As per experimental data
Dissociation constant of water at 25 ℃ is
10^ (-14)= [H+] [OH-]
So at 25℃ [ H+]= [OH-] =10^(-7 )
Thus PH comes 7 at 25℃
Dissociation of water is an endothermic process ,so on increasing temperature equilibrium constant of dissociation of water will increase with temperature ,as temperature is favourable condition .
So eq. Constant for H2O will be more than >10^(-14 )
Therefore
[H+] at 85℃ will more than 10^(-7)
PH = -log (H+)
So for sure PH will be less than 7 .
Answer: It must be less than 7 (<7)
Explanation: Actually adding NaCl doesn't affect the Ph, whatsoever the tempreature may be because NaCl is a salt of strong acid and strong base and it will produce equal amont of H+ and OH- when dissolved in water.
The actual reason for Ph<7 is tempreature. At 25 degree celcius dissociation constant of water that is,
Kw = 10^-14 and [H+] = 10^-7
But at higher tempreture, Kw>10^-14
.This also implies [H+] > 10^-7 so Ph is less than 7. So without NaCl also Ph of pure water at 85 degrees celcius is less than 7 (<7).