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I got that pH=3.65 using the fact that Ka=[H⁺][A⁻]/[HA] at equilibrium. In the ice table, I stands for initial, C stands for change, and E stands for equilibrium.
I hope this helps. Let me know if anything is unclear.
I hope this helps. Let me know if anything is unclear.
Missing data in your question: (please check the attached photo)
from this balanced equation:
M(OH)2(s) ↔ M2+(aq) + 2OH-(aq) and when we have Ksp = 2x10^-16
∴Ksp = [M2+][OH]^2
2x10^-16 = [M2+][OH]^2
a) SO at PH = 7
∴POH = 14-PH = 14- 7 = 7
when POH = -㏒[OH]
7= -㏒[OH]
∴[OH] = 1x10^-7 m by substitution with this value in the Ksp formula,
∴[M2+] =Ksp /[OH]^2
= (2x10^-16)/(1x10^-7)^2
= 0.02 M
b) at PH =10
when POH = 14- PH = 14-10 = 4
when POH = -㏒[OH-]
4 = -㏒[OH-]
∴[OH] = 1x10^-4 ,by substitution with this value in the Ksp formula
[M2+] = Ksp/ [OH]^2
= 2x10^-16 / (1x10^-4)^2
= 2x10^-8 M
c) at PH= 14
when POH = 14-PH
= 14 - 14
= 0
when POH = -㏒[OH]
0 = - ㏒[OH]
∴[OH] = 1 m
by substitution with this value in Ksp formula :
[M2+] = Ksp / [OH]^2
= (2x10^-16) / 1^2
= 2x10^-16 M
from this balanced equation:
M(OH)2(s) ↔ M2+(aq) + 2OH-(aq) and when we have Ksp = 2x10^-16
∴Ksp = [M2+][OH]^2
2x10^-16 = [M2+][OH]^2
a) SO at PH = 7
∴POH = 14-PH = 14- 7 = 7
when POH = -㏒[OH]
7= -㏒[OH]
∴[OH] = 1x10^-7 m by substitution with this value in the Ksp formula,
∴[M2+] =Ksp /[OH]^2
= (2x10^-16)/(1x10^-7)^2
= 0.02 M
b) at PH =10
when POH = 14- PH = 14-10 = 4
when POH = -㏒[OH-]
4 = -㏒[OH-]
∴[OH] = 1x10^-4 ,by substitution with this value in the Ksp formula
[M2+] = Ksp/ [OH]^2
= 2x10^-16 / (1x10^-4)^2
= 2x10^-8 M
c) at PH= 14
when POH = 14-PH
= 14 - 14
= 0
when POH = -㏒[OH]
0 = - ㏒[OH]
∴[OH] = 1 m
by substitution with this value in Ksp formula :
[M2+] = Ksp / [OH]^2
= (2x10^-16) / 1^2
= 2x10^-16 M