You can use the follwing equations.
pH=14-pOH
pOH=-log[OH⁻]
a) pOH=-log10⁻¹²
pOH=12
pH=14-12
pH=2
You can also use these equations:
[H⁺]=K(w)/[OH⁻] (K(w)=1.01×10⁻¹⁴)
pH=-log[H⁺]
b) [H⁺]=(1.01×10⁻¹⁴)/10⁻²M
[H⁺]=1.01×10⁻¹²M²
pH=-log(1.01×10⁻¹²)
pH=12
You can use either method. It does not really matter.
I hope this helps. Let me know if anything is unclear and when you do the calculations for c you should get pH=7.
