1)Using tabulated K a and K b values, calculate the pH of each of the following
ID: 491343 • Letter: 1
Question
1)Using tabulated Ka and Kb values, calculate the pH of each of the following solutions.
a.A solution containing 0.10 M HNO2 and 0.15 M NaNO2
b.25.0 g pure (glacial) acetic acid and 40.0 g sodium acetate diluted to 500. mL with water.
c.50.0 mL of 1.0 M HOCl and 30.0 mL of 0.80 M NaOH diluted to 250 mL with water,
d.100.0 g NH4Cl and 65.0 g NaOH diluted to 1.0 L with water.
e.26.4 g sodium acetate and 50.0 mL of 6.00 M hydrochloric acid diluted to 5.00 × 102 mL with water.
Hint: In parts c,d, and e, reactions occur that go to completion.
Explanation / Answer
1. pH= PKa+ log {[Conjugate base, NaNO2]/[Acid, HNO2]}
Ka of HNO2= 4*10-4, pKa= 3.398, pH= 3.398+log(0.15/0.1)= 3,574
2. molar masses : acetic acid (CH3COOH)= 60, sodium acetate (CH3COONa)= 82
moles= mass/molar mass
moles : acetic acid :25/60= 0.42, sodium acetate = 40/82= 0.49
Concentration = moles/Volume(L) : acetic acid =0.42/0.5 = 0.84, sodium acetate=0.49/0.5=0.98
pH= pKa+log[conjugate base ,sodium acetate]./[acid, acetic acid)
Pka of acetic acid = 4.75, pH= 4.75+log (0.98/0.84)= 4.82
3.
moles = Molarity* Volume (L)
Moles : HOCl = 1*50/1000L= 0.05, for NaOH, moles = 0.8*30/1000 =0.024, this has been diluted to 250ml with water
molarity of NaOH after dilution = 0.8*30/250 =0.096M
the reaction between HOCl and NaOH is HOCl+ NaOH--------->NaCl+H2O
molar ratio (theoretical) between HOCl and NaOH= 1:1
actual molar ratio = 0.05:0.024= 2.08 :1, excess is HOCl and is excess by = 0.05-0.024= 0.026
Volume after mixing = 50+250= 300ml=0.3L, concentration of HOCl= 0.026/0.3=0.087M
HOCl ionizes as HOCl + H2O -------->H3O+ OCl-
Ka= [H3O+] [OCl-]/[HOCl]
let x= drop in concentration of HOCl to reach equilibrium. At equilibrium [HOCl]=0.087-x, [H3O+] = [OCl-]=x
Ka for HOCl = 7.53, Ka= 10(-7.53)= 2.95*10-8 = x2/(0.087-x)= 2.95*10-8, when solved using excel, x= 5.05*10-5
pH= -log (5.05*10-5)= 4.29
100.0 g NH4Cl and 65.0 g NaOH diluted to 1.0 L with water.
4. moles of NH4Cl = 100/53.5 = 1.86, moles of NaOH= 65/40= 1.625, the reaction is
OH- + NH4+ ------>NH3+H2O
molar ratio of NH4Cl:NaOH= 1:1 actuak ratio given = 1.86 :1.625 = 1.14 :1. So excess is NH4Cl
excess NH4Cl= 1.86-1.625= 0.235, moles of NH3 formed=1.625
Ka for NH4Cl = 5.4*10-10
NH4+ + H2O -------->NH3 + H3O+
concentrations = NH4+ = 0.235/1=0.235, NH3= 1.625
Ka= [NH3+] [H3O+]/[NH4+]
5.4*10-10 = 1.625[H+]/0.235
[H+] = 5.4*10-10*0.235/1.625=7.81*10-11
pH= -log[H+]= 10.1
Ka= [NH3] [H+]/[NH4+]
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