Question

a) What volume (to the nearest 0.1 mL) of 6.60-M HCl must be added to 0.450 L of 0.200-M K₂HPO₄ to prepare a pH = 7.40 buffer?

b) What volume (to the nearest 0.1 mL) of 7.10-M NaOH must be added to 0.600 L of 0.250-M HNO₂ to prepare a pH = 3.10 buffer?

Answer 1

a) What volume (to the nearest 0.1 mL) of 6.60-M HCl must be added to 0.450 L of 0.200-M K2HPO4 to prepare a pH = 7.40 buffer?

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

initially

mol of HPO4-2 = 0.200*0.45 = 0.09

after adding H+ then

mol of HPO4-2 = 0.09 - x

mol of H2PO4- formed = 0 + x

subtitute in buffer equation

7.40 = 7.21 + log(HPO4-2 / H2PO4-)

7.40 = 7.21 + log(x / (0.09-x))

10^(7.40-7.21) = x / (0.09-x)

0.6456x = 0.09 - x

1.6456 x= 0.09

x = 0.09/1.6456 = 0.05469

then, we must add

V = mol/M = 0.05469/6.6 = 0.008286 L = 8.29 mL of acid required

b) What volume (to the nearest 0.1 mL) of 7.10-M NaOH must be added to 0.600 L of 0.250-M HNO2 to prepare a pH = 3.10 buffer?

pH = pKa + log(NO2-/HNO2)

3.10 = 3.39 + log(NO2-/HNO2)

10^(3.10-3.39) =  + log(NO2-/HNO2)

0.5128 =  (NO2-/HNO2)

initially

mol of HNO2 = MV = 0.6*0.25 = 0.15

after x mol of NaOH

mol of HNO2 = 0.15-x

mol of NO2- = x

0.5128 =  (NO2-/HNO2)

0.5128 =  x / (0.15-x)

0.15-x = 1.950x

(1+1.950)x= 0.15

x =(0.15) / (1.95+1) = 0.05084

M = mol/V

V = mol/M = (0.05084)/7.10

V = 0.00716 L = 7.16 mL