Question

Calculate the pKb of a weak base BOH, which gives a pH of 10.5 when 0.01 moles of the base is dissolved in 500 mL of water. Select one: O a. 3.5 O b. 1.7 O c. 2.0 O d. 3.16 o e. 5.3

Answer 1

atot Molar concentration or Molarity (M) is defined as the number of molen of volute dinsolve in one liten of not Therefore, Molari concentration = Number o - Literin of solution , Given, 0.01 moles of the borse in dianalved in 500 mL of Then, number of mole of base (nolute) = 0.01 molen volume of the solution = 500 ml = 500 OL 1000 0.01 mol 14- = 0.5 L Therefore, molari concentration of the solution = a. 0.5L. = 0.02 mol/L = 0.02 M iven, pH of the solution = 10.5 Now, for a weak bare, Boh pOH = 1/₂ pkb _I Logo = 14-pH = 1 pkb _ / Loge [Ac at 25 CM7 = 14-10.5 = 1PKD -!Log(0.02) → 3.5 = {pkb - '*(-1,699) → 3.5 = 'PKO + 0.8495 ► lapkb = 3.5 -0.8495 pkb = 2.6505 + pkb = 2 x 2 6 505 → pkb = 5.3 Therefore, the pkb of the weak bare, Bott is 5.3. 2 7