Answer
A) 12.7
B) 1
C) 5.22
Work Step by Step
a) Concentration of NaOH in 1 L solution = 0.01 L* 5.0 M/1.0 L = 0.05 M 0.05M NaOH is a strong base. It completely gets ionised togive OH ions of 0.05M. pOH = -log [OH-] = -log(0.05) = 1.30 pH = 12.7
b) Glycine is a weak acid and its dissociation is supressed bya strong acid HCI. .. pH of the solution only depends upon concentration ofHCI. Being a strong acid it completely gets = 0.1 pH = -log [H+] ionised to giveH+ ions and Cl ions. [H+] = 0.02 L * 5.0 M / 1.00 L Number of moles of aceticacid = 0.01 L * = 1
c) Number moles of sodium acetate = = 5 g/ 82.03 gmol-1 = 0.06 moles 2.0 M =0.02 moles pka = 4.75 Formula: pH = pka + log [salt] / [acid] As the volume is same, concentrationcan be replaced by number of moles. pH = 4.75 + log (0.06 / 0.02) = 5.22
