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.
