Question

20. If 10 mL of 0.05 M NaOH is added to a 20 mL solution of 0.1 M NaNO2 and 0.1 M HNO2, what will be the pH of the resultant solution? Assume that volumes are additive. Ka for HNO2 = 7.1x10-4. 21. At 25°C, 50.0 mL of 0.50 M NaOH(aq) is added to a 250 mL aqueous solution containing 0.30 M NH3 and 0.36 M NH4Cl. What is the pH of the solution after the addition of the base? Assume that volumes are additive. Kb of NH3 = 1.8 x 10-5 22. Given 1.00 L of a solution that is 0.100 M in sodium propionate (NaC3H5Oz) and 0.300 M in propionic acid (HC3H5O2), what is the pH after 0.0400 mole of HNO3 is added? Assume that the volume does not -change upon addition of the HNO3. Ka for HC3H5O2 = 1.3 x 10-5

Answer 1

20)

HNO2 +NaOH ------------------> NaNO2 + H2O

20x0.1= 2 0 20x0.1=2 ----------- initial mmoles

---------- 10x0.05 = 0.5 -------------- ----------- change

1.5 0 2.5 --------- equilibrium

pKa of HNO2 = -log 7.1x10-4  = 3.15

The pH of this buffer = pKa + log [conjugate base]/[acid]

= 3.15 + log 2.5/1.5

= 3.37

21)

NH4Cl + NaOH ------------------> NH3 + NaCl + H2O

250x0.36 =90 0 250x0.3= 75 ----------------------- initial mmoles

----------- 50x0.5= 25 ---------- ----------------------- change

65 0 100 ------------ equilibrium

The pKb of NH3 = -log 1.8 x10-5 = 4.74

pOH of the base is given by Hendersen equation as

pOH = pKb + log [conjugate acid]/[base]

= 4.74 + log 65/100 = 4.55

and pH = 14-pOH = 14-4.55=9.45