(a) 0.100 M HCl    [H⁺] = 0.100 M    pH = 1.00

(b) HCl:     mol = n = MxL = (0.100)(0.0400) = 0.00400 mol
     NaOH:  n = MxL = (0.100)(0.0200) = 0.00200 mol
     HCl in excess by 0.00200 mol,  volume = 60.00 mL = 0.06000 L
     [H⁺] = 0.00200mol/0.06000L = 0.0333 M      pH = 1.48

(c) HCl:     n = MxL = (0.100)(0.0400) = 0.00400 mol
     NaOH:  n = MxL = (0.100)(0.0300) = 0.00300 mol
     HCl in excess by 0.00100 mol,  volume = 70.00 mL = 0.07000 L
     [H⁺] = 0.00100mol/0.07000L = 0.0143 M      pH = 1.84

(d) HCl:     n = MxL = (0.100)(0.0400) = 0.00400 mol
     NaOH:  n = MxL = (0.100)(0.0400) = 0.00400 mol
     Solution neutral   pH = 7.00

(e) HCl:     n = MxL = (0.100)(0.0400) = 0.00400 mol
     NaOH:  n = MxL = (0.100)(0.0500) = 0.00500 mol
     NaOH in excess by 0.00100 mol,  volume = 90.00 mL = 0.09000 L
     [OH^-1] = 0.00100mol/0.09000L = 0.0111 M   pOH = 1.95  pH= 12.05

Sample Problem 2:
What is pH if 50.0 mL of 0.200 M HCl mixed with 30.0 mL of 0.150 M NaOH?
HCl:      n = MxL = (0.200)(0.0500) = 0.0100 mol
NaOH:  n = MxL = (0.150)(0.0300) = 0.00450 mol
HCl in excess by 0.00550 mol,  volume = 80.0 mL = 0.0800 L
[H⁺] = 0.00550 mol/0.0800L = 0.0688 M      pH = 1.16

Sample Problem 3:
What is pH if 35.0 mL of 0.125 M HCl mixed with 45.0 mL of 0.150 M NaOH?
HCl:      n = MxL = (0.125)(0.0350) = 0.00438 mol
NaOH:  n = MxL = (0.150)(0.0450) = 0.00675 mol
NaOH in excess by 0.00237 mol,  volume = 80.0 mL = 0.0800 L
[OH^-1] = 0.00237 mol/0.08000L = 0.0296 M      pOH = 1.53   pH = 12.47