Question

We ariswer UHUILE la vest cumpeles lile Sidlenen UI answers me Yuugu . Consider the titration of 300.0 mL of 0.544 M NHz (Kb = 1.8 x 105) with 0.500 M HNO3. After 150.0 mL of 0.500 M HNO3 have been added, the pH of the solution is: Ca. 4.67 b. 11.33 c. 9.33 d. 6.33 e. none of these

Answer 1

pKb = - log (1.8 * 10^-5) = 4.745

mole of NH3 = 0.300 L * 0.544 mole / L = 0.1632 mole.

and

mole of HNO3 = 0.150 L * 0.500 mole / L = 0.075 mole.

thus

mole of salt = 0.075 mole

and

mole of base (NH3) = 0.1632 - 0.075 = 0.0882 mole.

the resulting solution is basic buffer solution.

pOH = pKb + log [salt] / [base]

or

pOH = 4.745 + log (0.075 / 0.0882)

or

pOH = 4.67

pH + pOH = 14

or

pH = 14 - 4.67

or

pH = 9.33

option (c) 9.33 is the answer.