Need help with these questions please

Question

1. Calculate the [OH-] and the pH of a solution with [H+]=3.3 x 10^-9 M at 25 C
2. calculate the [H+] and the pOH of a solution with [OH-]=1.7 x 10^-12 M at 25 C
3. calculate the [H+] and [OH-] of a solution with pH=7.3 at 25 C

Answer 1

Hello Shenelle,

For q1: pH = - log [H+], and since [H+] = 3.3 x 10^-9 M,

so the pH = - log (3.3 x 10^-9) = 8.48

Also, Kw = [OH-] [H+]. So, [OH-] = Kw / [H+]. Kw = 1 x 10^-14.

So, [OH-] = (1 x 10^-14) / (3.3 x 10^-9) = 3.0 x 10^-6 M

For q 2: pOH = - log [OH-], and since [OH-] = 1.7 x 10^-12 M

so pOH = - log (1.7 x 10^-12 M) = 11.77

Also, [H+] = Kw / [OH-], and Kw = 1 x 10^-14

So, [H+] = (1 x 10^-14) / (1.7 x 10^-12) = 5.9 x 10^-3

For q3: The formula for pH = - log [H+], and pH = 7.3

According to the rules of log, log x = y is x = 10^y

So, [H+] = 10^-pH,

[H+] = 10^-7.3 = 5 x 10^-8

[OH-] = Kw / [H+] = (1 x 10^-14) / (5 x 10^-8) = 2 x 10^-7