Question

calculate the {H3O+} and {OH-} for a solution with the following oH values:
{H3O},{OH-} = 
(separate by a comma on each)

1. pH= 10.3 (one significant figured)

2. pH=6.0 (one Sig fig)

3. pH= 7.35 (two sig fig)

4. pH=6.1 (one Sig fig)

5. pH= 1.24 (two sig fig) 

Answer 1

note that

Kw = [H3O+][OH-] = 10^-14

pOH + pH = 14

so

1)

pH = 10.3

pOH = 14-10.3 = 3.7

[OH-] = 10^-pOH = 10^-3.7 = 0.000199 = 2*10^- 4

[H3O+] = 10^-pH = 10^-10.3 = 5.01*10^-11 = 5*10^-11

2

pH = 6

pOH = 14-6= 8

[OH-] = 10^-pOH = 10^-8

[H3O+] = 10^-pH = 10^-6

Q3.

pH = 7.35

pOH = 14-7.35 = 6.65

[OH-] = 10^-pOH = 10^-6.65 = 2.2*10^-7

[H3O+] = 10^-pH = 10^-7.35 = 4.5*10^-8

Q4

pH = 6.1

pOH = 14-6.1= 7.9

[OH-] = 10^-pOH = 10^-7.9= 1.3*10^-8

[H3O+] = 10^-pH = 10^-6.1= 7.9*10^-7

q5)

pH = 1.24

pOH = 14-1.24 = 12.76

[OH-] = 10^-pOH = 10^-12.76= 0.0575 = 5.8*10^-2

[H3O+] = 10^-pH = 10^-1.24=1.7*10^-13