Question

1

a. How many grams (to the nearest 0.01 g) of NH4Cl (Mm = 53.49 g/mol) must be added to 900. mL of 0.923-M solution of NH3 in order to prepare a pH = 8.90 buffer?

b. What volume (to the nearest 0.1 mL) of 6.70-M NaOH must be added to 0.550 L of 0.250-M HNO₂ to prepare a pH = 3.00 buffer

c. What volume (to the nearest 0.1 mL) of 6.20-M HCl must be added to 0.400 L of 0.200-M K₂HPO₄ to prepare a pH = 7.30 buffer?

Answer 1

a)

pH = pKa + log(NH3/NH4+)

pKa = 9.25 for NH4+

then

8.90 = 9.25 + log(NH3/(NH4+))

8.90 = 9.25 + log((0.923) / [NH4+])

[NH4+] = (10^-(8.9-9.25)) / (0.923)

[NH4+] = 2.4254 M

mol of NH4+ = MV = 2.4254*0.900 = 2.18286

mass = mol*MW = 2.18286*53.49 = 116.76 g of NH4Cl

b)

pH = pKa + log(NO2-/HNO2)

mol = MV = (0.55)(0.25) = 0.1375 mol

3 = 3.32 + log( x /(0.1375-x))

10^(3-3.2) = x /(0.1375-x)

1.5849x = (0.1375-x)

2.5849x = 0.1375

x = 0.1375/2.5849

x = 0.05319 mmol of H+ required

V = mol/M = 0.05319 / 6.7 = 0.0079 L = 7.94 mL

c)

K2HPO4 = HPO4-

HPO4- + HCl = H2PO4

then

pH = pKa + log(HPO4-2 / H2PO4-)

mol = MV = (0.4*0.2) = 0.08

7.30 = 7.21 + log( (0.08 - x) / x)

10^(7.30 -7.21 ) =  (0.08 - x) / x

1.2302x = 0.08-x

2.2302x = 0.08

x = 0.08/2.2302

x = 0.03587 mol

V = mol/M = 0.03587/6.2 = 0.005785L = 0.005785*10^3 = 5.785 mL