Question

You have the following solutions at your disposal to prepare a buffer solution with a pH greater than 7.0:

(1) 50.0 mL of 0.35 M NH3 Kb = 1.8×10-5  (2) 55.0 mL of 0.35 M NaOH

(3) 50.0 mL of 0.35 M HCl (4) 16.5 mL of 0.35 M NaCl NaOH

(5) 16.5 mL of 0.35 M HNO3 (6) 16.5 mL of 0.35 M KOH

(a) Assuming that you are going to mix the entire quantity of the listed solutions to prepare the buffer, which two solutions would you use?

Answer 1

buffer is weak base + salt of its strong acid

so we need to check combination (1,3) or (1,5)

(1) 50.0 mL of 0.35 M NH3 :

millimoles of NH3 = 50 x 0.35 = 17.5

HCl millimoles = 50 x 0.35 = 17.5

NH3 + HCl -------------------> NH4Cl

17.5     17.5                              0          ---------------> initial

0           0                                17.5 --------------------> after reaction

NH4Cl is the salt of weak base strong acid

pH < 7

so now check for (1,5)

millimoles of HNO3 = 16.5 x 0.35 = 5.78

NH3 + HNO3 --------------------> NH4NO3

17.5     5.78                                0               ---------------> initial

11.72      0                               5.78

so NH3 + NH4NO3 is a buffer

pOH = pKb + log [NH4NO3/NH3]

pOH = 4.75 + log (5.78 /11.72)

pOH = 4.44

pH + pOH =14

pH = 9.56

so

solution 1+ solution 5 should be used