What is the ph in a solution of acid containing [h+] = 5x10-9 m?
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f we use the relation, pH = – log [H3O+], we get pH equal to 8. But this is not correct because an acidic solution cannot have pH greater than 7. It may be noted that in very dilute acidic solution, when H+concentrations from acid and water are comparable, the concentration of H+ from water cannot be neglected.
Therefore,
[H+] total = [H+] acid + [H+] water
Since HCl is a strong acid and is completely ionized
[H+] HCl = 1.0 x 10-8
The concentration of H+ from ionization is equal to the [OH–] from water,
[H+] H2O = [OH–] H2O
= x (say)
[H+] total = 1.0 x 10-8 + x
But
[H+] [OH–] = 1.0 x 10-14
(1.0 x 10-8 + x) (x) = 1.0 x 10-14
X2 + 10-8 x – 10-14 = 0
Solving for x, we get x = 9.5 x 10-8
Therefore,
[H+] = 1.0 x 10-8 + 9.5 x 10-8
= 10.5 x 10-8
= 1.05 x 10-7
pH = – log [H+] = – log (1.05 x 10-7) = 6.98
Therefore,
[H+] total = [H+] acid + [H+] water
Since HCl is a strong acid and is completely ionized
[H+] HCl = 1.0 x 10-8
The concentration of H+ from ionization is equal to the [OH–] from water,
[H+] H2O = [OH–] H2O
= x (say)
[H+] total = 1.0 x 10-8 + x
But
[H+] [OH–] = 1.0 x 10-14
(1.0 x 10-8 + x) (x) = 1.0 x 10-14
X2 + 10-8 x – 10-14 = 0
Solving for x, we get x = 9.5 x 10-8
Therefore,
[H+] = 1.0 x 10-8 + 9.5 x 10-8
= 10.5 x 10-8
= 1.05 x 10-7
pH = – log [H+] = – log (1.05 x 10-7) = 6.98
Answered by
1
ph"= 6.98
And remember, the pH of an acidic solution can never, ever, be higher than 7!
And remember, the pH of an acidic solution can never, ever, be higher than 7!
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