Question

a.) What volume (to the nearest 0.1 mL) of 4.50-M NaOH must be added to 0.300 L of 0.150-M HNO2 to prepare a pH = 3.20 buffer? ______ mL

b.) What volume (to the nearest 0.1 mL) of 4.00-M HCl must be added to 0.700 L of 0.350-M K₂HPO₄ to prepare a pH = 7.50 buffer? _______ mL

Answer 1

For a buffer, we require NO2- and HNO2

so

pKa of HNO2 = 3.39

so

the buffer equation (henderson hasselbach)

pH = pKa + log(NO2-/HNO2)

3.20 = 3.39 + log(NO2-/HNO2)

initially:

mol of HNO2 = V = 0.3*0.15 = 0.045

mol of NO2- = 0

after adding --> mol of base = M*Vbase = 4.5*Vbase

mol of HNO2 = 0.045 - 4.5*Vbase

mol of NO2- = 0 + 4.5*Vbase

substitte

3.20 = 3.39 + log(NO2-/HNO2)

3.20 = 3.39 + log((4.5*Vbase)/( 0.045 - 4.5*Vbase) )

10^(3.20-3.39) = 4.5*Vbase/( 0.045 - 4.5*Vbase)

0.6456/4.5*(0.045 ) + 0.6456/4.5*4.5*Vbase = Vbase

Vbase = (0.6456/4.5*(0.045 ) )/(1- 0.6456) = 0.01821liters = 0.01821 *10^3 mL = 18.21 mL

b)

apply same logic:

pH is in the range of the second ionization so, we need:

H2PO4- and HPO4-

K2HPO4- --> 2K+ + HPO4-

HPO4- + H+ --> H2PO4-

so..

initially

HPO4-2 = MV = 0.7*0.35 = 0.245 mol

H2PO4- = 0

after adding mol of HCl = Macid*Vacid = 4*Vacid

HPO4-2 = 0.245 - 4*Vacid

H2PO4- = 0 + 4*Vacid

from the buffer equation:

pKa2 = 7.21

pH = pKa2 + log(HPO4-2 / H2PO4- )

7.50 = 7.21 + log((0.245 - 4*Vacid ) /( 4*Vacid))

solve for Vacid

10^(7.50-7.21) = (0.245 - 4*Vacid ) /( 4*Vacid))

1.9498*4*Vacid = 0.245 - 4*Vacid

0.245 = (1.9498*4+4)*Vacid

Vacid = 0.245 / (1.9498*4+4) = 0.020764 L = 0.020764*10^3 mL= 20.764 mL of acid