Homework Answers
5)
Given:
Ka = 4.9*10^-10
use:
Kb = Kw/Ka
Kw is dissociation constant of water whose value is 1.0*10^-14 at 25 oC
Kb = (1.0*10^-14)/Ka
Kb = (1.0*10^-14)/4.9*10^-10
Kb = 2.041*10^-5
Answer: 2.0*10^-5
6)
use:
pH = -log [H+]
2.8 = -log [H+]
[H+] = 1.585*10^-3 M
HA dissociates as:
HA -----> H+ + A-
0.294 0 0
0.294-x x x
Ka = [H+][A-]/[HA]
Ka = x*x/(c-x)
Ka = 1.585*10^-3*1.585*10^-3/(0.294-1.585*10^-3)
Ka = 8.59*10^-6
Answer: 8.59*10^-6
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5. Determine the Ky for CN at 25°C. The K, for...
CHM2046L: General Chemistry and Oualitative Analysis II; Fall 2019 Equilibrium: Determination of Acid Ionization Constant (Ka)...
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Question 1 7.5 pts Determine the Ka of a weak monoprotic acid with an initial concentration...
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a) b) c) d) If the Kb of a weak acid is 2.9 10*, the K,...
a)
b)
c)
d)
If the Kb of a weak acid is 2.9 10*, the K, of its conjugate acid must be which of the following? 2.9.10 6.46 7.54 3.4 10- The hydronium ion concentration of a 0.50 mol/L solution of hydrocyanic acid, HCN(aq), is 1.4 x 10-5 mol/L. Use Weak Acid Info 16.1 to answer the following question. The Ka for hydrocyanic acid is: 3.9 x 10-10 1.4 x 10-5 5.0 x 10-2 6.2 x 10-10 An acetic acid-sodium...
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(NaOH= 0.5 M). Determine the pH before any base has been added.
Please show steps and please explain why the answer is what it
is.
24B) The 40mL 0.2 M formic acid is titrated with a 6.0mL of
strong base. Here NaOH can be treated as a conjugate base and
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please show steps and explain!
Us (24-25, Acid/base, aqueous equilibrium) 24A) (4...
The original given concentration was 25.0mL of 0.100 M HCO2H (formic acid Ka= 1.8x10^-4) with 0.100...
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4. 16.55 Calculate the pH of an aqueous solution at 25°C that is 0.095 M in...
4. 16.55 Calculate the pH of an aqueous solution at 25°C that is 0.095 M in hydrocyanic acid (HCN). (K for hydrocyanic acid=4.9 x 1010 Answers TOmTzctu. 10. 16.61 Calculate the Ka of a weak acid if a 0.19-M aqueous solution of the acid has a pH of 4.52 at 25°C Answers 2. 16.71 The pH of a 0.30-M solution of a weak base is 10.66 at 25°C. What is the K of the base? Answers 4. 16.73 Calculate the...
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TABLE 15.5 Acid lonization Constants (K) for Some Monoprotic Weak Acids at 25 °C Acid Formula Structural Formula lonization Reaction К. pk, = -log (2) Chlorous acid HCIO H-0-C=0 1.1 x 10-2 1.96 Stronger acids Nitrous acid HNOZ H-O-NO 5.6 x 10-4 HCIO2(aq) + H2011) = H30* (aq) + CO2 (aq) HNO2(aq) + H20(I) = H30+ (aq) + NO2 (aq) HF(aq) + H20(1) = H30* (aq) + F"(aq) 3.25 Hydrofluoric acid HF H-F 6.3 x 10-4 3.20 O Methanoic acid...
Question 1 7.5 pts Determine the Ka of a weak monoprotic acid with an initial concentration of 0.294 M and has a pH of 2.80 once equilibrium is reached. O 1.2* 10.5 8.6 x 106 O 4.9 × 107 0 5.4× 103 2.7
a)
b)
c)
d)
If the Kb of a weak acid is 2.9 10*, the K, of its conjugate acid must be which of the following? 2.9.10 6.46 7.54 3.4 10- The hydronium ion concentration of a 0.50 mol/L solution of hydrocyanic acid, HCN(aq), is 1.4 x 10-5 mol/L. Use Weak Acid Info 16.1 to answer the following question. The Ka for hydrocyanic acid is: 3.9 x 10-10 1.4 x 10-5 5.0 x 10-2 6.2 x 10-10 An acetic acid-sodium...
please explain. thanks
5. Find the pH and fraction of dissociation for a 0.0SF solution of the weak monoprotic acid HA whose Ka=1.4 x 106. Find the equilibrium molar concentrations of HA, A', H20 and OH, all species, for the solution. Make simplyfing assumptions to avoid solving a quadratic equation State rational for these assumptions
CHM2046L: General Chemistry and Qualitative Analysis II; Fall 2019 Lab # 9: EQUILIBRIUM: Calculating pH and Buffer Capacity Post-lab # 9: ( Due Monday, November 7t, 2019) Last name First Name 1. Define buffer capacity. 2. Determine the Kb for CN at 25°C. The Ka for HCN is 4.9 x 10-10, 3. Determine the [H30] in a 0.265 M HCIO solution. The Ka of HCIO is 2.9 x 10-8 rco 4. Determine the pH of a 0.461 M C6H5CO2H M...
24A) 40mL of 0.2 M formic acid is titrated with a strong base
(NaOH= 0.5 M). Determine the pH before any base has been added.
Please show steps and please explain why the answer is what it
is.
24B) The 40mL 0.2 M formic acid is titrated with a 6.0mL of
strong base. Here NaOH can be treated as a conjugate base and
formic acid is the acid.
please show steps and explain!
Us (24-25, Acid/base, aqueous equilibrium) 24A) (4...
The original given concentration was 25.0mL of 0.100 M HCO2H
(formic acid Ka= 1.8x10^-4) with 0.100 M of NaOH.
f) pH after adding 25.0 mL NaOH (Equivalence point). At this point 0.00250" ist 00250 mol of OH have been added, and therefore all the acid (HCO2H) has been converted into its conjugale the table below. 25.0 mL x L/1000 mL = 0.025 L 0.025 L x 0.1 mol/L = 0.0025 mol of NaOH added n converted into its conjugate base...
4. 16.55 Calculate the pH of an aqueous solution at 25°C that is 0.095 M in hydrocyanic acid (HCN). (K for hydrocyanic acid=4.9 x 1010 Answers TOmTzctu. 10. 16.61 Calculate the Ka of a weak acid if a 0.19-M aqueous solution of the acid has a pH of 4.52 at 25°C Answers 2. 16.71 The pH of a 0.30-M solution of a weak base is 10.66 at 25°C. What is the K of the base? Answers 4. 16.73 Calculate the...
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