Question

What volume (to the nearest 0.1 mL) of 5.20-M NaOH must be added to 0.750 L of 0.300-M HNO2 to prepare a pH = 4.20 buffer?

ml

Answer 1

HNO2 <-> H+ + NO2-

so..

we must produce some NO-2

pKa for HNO2 = 3.39

so

pH = pKa + loG(NO2-/HNO2)

4.2 = 3.39 + log(NO2-/HNO2)

ratio = 10^(4.2-3.39) = 6.4565

NO2- = 6.4565 * HNO2

note...

we start with

mol of HNO2 = MV = 0.3*0.75 = 0.225 mol of HNO2

after adding

mol of NaOH = MV = 5.2*V

then

we must have:

mol of acid left = 0.225 - 5.2*V

mol of conjugate formed = 0 + 5.2V

and we know that

NO2- = 6.4565 * HNO2

so

5.2V = 6.4565 * ( 0.225 - 5.2*V)

solve for V

5.2V = 6.4565 *0.225 - 5.2*6.4565 *V

(5.2+33.5738)*V = 1.4527125

V = 1.4527125/(5.2+33.5738)

V = 0.03746 liters

mL = 0.03746*1000 = 37.46 mL of NaOH required