Question

What volume (to the nearest 0.1 mL) of 5.30-M NaOH must be added to 0.350 L of 0.350-M HNO₂ to prepare a pH = 3.20 buffer?

Answer 1

First, identify the buffer

HNO2 = weak acid

so

NO2- = conjugate base

therefore

the buffer equation Hendersn hasslebach

pH = pKa + log(NO2-/HNO2)

pKa for acid = 3.39

so

for pH = 320

3.20 = 3.39 + log(NO2-/HNO2)

initially

mol o fHNO2 = MV = 0.35*0.35 = 0.1225 mol

mol of NO2- = 0

after adding --> mol of base = Mbase*Vbase = 5.3*Vbase:

HNO2 reacts to form NO2-

mol of HNO2 left = 0.1225 - 5.3*Vbase

mol of NO2- = 0 +5.3*Vbase

substitute in pH

3.20 = 3.39 + log(NO2-/HNO2)

3.20 = 3.39 + log((5.3*Vbase)/(0.1225 - 5.3*Vbase)

solve fr Vbase

10^(3.20-3.39) = (5.3*Vbase)/(0.1225 - 5.3*Vbase)

0.64565 * 0.1225 - 5.3*0.64565 *Vbase = 5.3*Vbase

Vbase (5.3 +  5.3*0.64565) = 0.64565 * 0.1225

Vbase = (0.64565 * 0.1225) / (5.3 +  5.3*0.64565) = 0.00906817 liters

Vbase = 0.00906817*10^3 mL = 9.06817 mL