What is the pH of the solution with the buffer and after adding NaOH?

Ka = 1.82x10^-5, pKa = 4.74
pH = pKa + log[Ac-]/[HA]
[Ac-] = [HA]

HAc H+ + A-
A- + H+ (from HCl) HAc
moles Ac- = moles H+ = 1M x 1L = 1mole
moles HAc = 1moleH+ + 0.01moles H+(from HCl) = 1.01moles
[HAc] = 1.01moles / 1L = 1.01M
moles Ac- = 1mole - 0.01moles H+ = 0.99moles
[Ac-] = 0.99moles / 1L = 0.99M
pH = pKa + log[A-] / [HAc]
pH = 4.74 + log(0.99 / 1.01) = 4.73

the addition of 0.01moles OH- will remove 0.01moles H+ from HAc
moles HAc = 1 - 0.01 = 0.99moles, 0.99moles / 1L = 0.99M
moles Ac- = 0.01moles 1.01 / 1L = 1.01L
pH = 4.74 + log[1.01] / [0.99]
pH = 4.75