know the answers looking for how to work them out ! Review 45. Consider the titr
ID: 706461 • Letter: K
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know the answers looking for how to work them out !
Review 45. Consider the titration of 40.0 ml. of 0.200 M HCIO4 with 0.100 M KOH. Calculate the pll of the resulting solutions after the following volumes of KOH have been added. A) 0.0 mL B) 50.0 ml C) 40.0 mL D) 80.0 mL E) 100.0 ml 46. Consider the titration of 100 ml, of 0.20 M acetic acid (Ka 1.8 x 10-5) with 0.100 M KOH. Calculate the pH of the resulting solutions after the following volumes of KOH have been added. A) 0.0 ml B) 50.0 mL C) 100.0 mL D) 150.0 mL E) 200.0 mL F) 250.0 ml 47. A 65.0-mL sample of 0.12 MHNO, (K-4.0x 10-) is titrated with 0.11 M NaOH. What is the pH after 28.4 mL of NaOH has been added? A) 10.43 B) 7.00 C) 3.57 D) 3.00 E) 3.22 48. The value of Ksp for Agl is 1.5 x 10-16. Calculate the equilibrium concentration of Ag' in a 0.44 M Nal solution. A) 1.2× 10-8 mol/L B) 3.4 x 10 16 mol/L C) 2.7 x 109 mol/I D) 6.6x10-17mol/L E) 1.8 x 10 mol/L 49. Consider the titration of 100.0 ml. of 0.20 M acetic acid (Ka 1.8 x 10) with 0.10 M sodium hydroxide. Calculate the pH after 200 ml. of sodium hydroxide is added. A) 2.96 B) 8.79 C) 10.65 D) 5.21 E) none of the aboveExplanation / Answer
45. Titration
strong acid HClO4 with strong base KOH
HClO4 + KOH --> KClO4 + H2O
A) [HClO4] = 0.2 M
pH = -log[H+] = -log(0.2) = 0.7
B) HClO4 present = 0.2 M x 40 ml = 8 mmol
KOH added = 0.1 M x 50 ml = 5 mmol
excess HClO4 = 3 mmol/90 ml = 0.033 M
pH = -log(0.033) = 1.48
C) HClO4 present = 0.2 M x 40 ml = 8 mmol
KOH added = 0.1 M x 80 ml = 8 mmol
equivalence point
pH = 7
D) HClO4 present = 0.2 M x 40 ml = 8 mmol
KOH added = 0.1 M x 100 ml = 10 mmol
excess KOH present = 2 mmol/140 ml = 0.0143 M
pOH = -log[OH-] = -log(0.0143) = 1.84
pH = 14 - pOH = 14 - 1.84 = 12.15
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46. Titration of weak acid (CH3COOH) with strong base (KOH)
CH3COOH + KOH --> CH3COOK + H2O
A) [CH3COOH] = 0.2 M
CH3COOH + H2O <==> CH3COO- + H3O+
let x amount dissociated
Ka = [CH3COO-][H3O+]/[CH3COOH]
1.8 x 10^-5 = x^2/0.2
x = [H3O+] = 1.7 x 10^-3 M
pH = -log[H3O+] = -log(1.7 x 10^-3) = 2.72
B) CH3COOH present = 0.2 M x 100 ml = 20 mmol
KOH added = 0.1 M x 50 ml = 5 mmol
[CH3COONa] formed = 5 mmol/150 ml = 0.033 M
[CH3COOH] remained = 15 mmol/150 ml = 0.1 M
Using Hendersen-Hasselbalck equation,
pH = pKa + log(base/acid)
= 4.74 + log(0.033/0.1)
= 4.26
C) CH3COOH present = 0.2 M x 100 ml = 20 mmol
KOH added = 0.1 M x 100 ml = 10 mmol
[CH3COONa] formed = 10 mmol/200 ml = 0.05 M
[CH3COOH] remained = 10 mmol/200 ml = 0.05 M
Half-equivalence point
Using Hendersen-Hasselbalck equation,
pH = pKa + log(base/acid)
= 4.74 + log(0.05/0.05)
= 4.74
D) CH3COOH present = 0.2 M x 100 ml = 20 mmol
KOH added = 0.1 M x 150 ml = 15 mmol
[CH3COONa] formed = 15 mmol/250 ml = 0.06 M
[CH3COOH] remained = 5 mmol/250 ml = 0.02 M
Using Hendersen-Hasselbalck equation,
pH = pKa + log(base/acid)
= 4.74 + log(0.06/0.02)
= 5.22
E) CH3COOH present = 0.2 M x 100 ml = 20 mmol
KOH added = 0.1 M x 200 ml = 20 mmol
Equivalence point
[CH3COONa] formed = 20 mmol/300 ml = 0.067 M
CH3COONa --> CH3COO- + Na+
CH3COO- + H2O <==> CH3COOH + OH-
let x amount hydrolyzed
Kb = Kw/Ka = [CH3COOH][OH-]/[CH3COO-]
1 x 10^-14/1.8 x 10^-5 = x^2/0.067
x = [OH-] = 6.1 x 10^-6 M
pOH = -log[OH-] = 5.21
pH = 14 - pOH = 8.78
F) CH3COOH present = 0.2 M x 100 ml = 20 mmol
KOH added = 0.1 M x 250 ml = 25 mmol
excess [KOH] = [OH-] = 5 mmol/350 ml = 0.0143 M
pOH = -log[OH-] = 1.84
pH = 14 - pOH = 12.15
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47. HNO2 (weak acid) + NaOH (strong base) --> NaNO2 + H2O
HNO2 present = 0.12 M x 65 ml = 7.8 mmol
NaOH added = 0.11 M x 28.4 ml = 3.124 mmol
[NaNO2] formed = 3.124 mmol/93.4 ml = 0.033 M
[HNO2] remained = 4.676 mmol/93.4 ml = 0.05 M
Using Hendersen-Hasselbalck equation,
pH = pKa + log(base/acid)
= 3.4 + log(0.033/0.05)
= 3.21
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