Question

A strong acid will dissociate completely in water. Sulfuric acid is a strong acid. H2SO4 (l) + 2H2O (l) -> 2H3O+ (aq) + SO42- (aq) Ka (sulfuric acid) = A) Based on this definition, if you dissolved a 2X10-1 M sulfuric acid in water, the concentration of hydronium coming only from the acid alone is: Show calculation using the appropriate mol:mol ratio from the above dissociation reaction. B) The solvent, water itself undergoes autodissociation/autoproteolysis and the equilibrium constant for that process (shown below), Kw, is 10-14 at 25 °C. Based on the dissociation reaction shown below and the Kw value provided, what is the concentration of hydronium coming ONLY from the solvent, water? Show calculation. C) Compare the hydronium concentrations calculated in questions 1A and 1B and identify which one is bigger and why? D) Calculate the pH of a solution containing 2X10-1 M sulfuric acid in water considering the hydronium generated from the acid as well as the solvent. Where pH = -log[H3O+]. E) Calculate the pH of the solution containing 2X10-1 M sulfuric acid in water considering the hydronium generated ONLY from the acid. Where pH = -log[H3O+]. F) Most common pH meters can only read two decimal places. Based on this information, would you, for all practical purposes, consider the hydronium concentration coming from water when considering the pH of a strong acid? Explain. 

          A strong acid will dissociate completely in water. Sulfuric acid is a strong acid.
H2SO4 (l) + 2H2O (l) -> 2H3O+ (aq) + SO42- (aq) Ka (sulfuric acid) =
A) Based on this definition, if you dissolved a 2X10-1 M sulfuric acid in water, the concentration of hydronium coming only from the acid alone is: Show calculation using the appropriate mol:mol ratio from the above dissociation reaction.
B) The solvent, water itself undergoes autodissociation/autoproteolysis and the equilibrium constant for that process (shown below), Kw, is 10-14 at 25 °C. Based on the dissociation reaction shown below and the Kw value provided, what is the concentration of hydronium coming ONLY from the solvent, water? Show calculation.
C) Compare the hydronium concentrations calculated in questions 1A and 1B and identify which one is bigger and why?
D) Calculate the pH of a solution containing 2X10-1 M sulfuric acid in water considering the hydronium generated from the acid as well as the solvent. Where pH = -log[H3O+].
E) Calculate the pH of the solution containing 2X10-1 M sulfuric acid in water considering the hydronium generated ONLY from the acid. Where pH = -log[H3O+].
F) Most common pH meters can only read two decimal places. Based on this information, would you, for all practical purposes, consider the hydronium concentration coming from water when considering the pH of a strong acid? Explain.
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Chemistry: Structure and Properties
Chemistry: Structure and Properties
Nivaldo Tro 2nd Edition
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A strong acid will dissociate completely in water. Sulfuric acid is a strong acid. H2SO4 (l) + 2H2O (l) -> 2H3O+ (aq) + SO42- (aq) Ka (sulfuric acid) = A) Based on this definition, if you dissolved a 2X10-1 M sulfuric acid in water, the concentration of hydronium coming only from the acid alone is: Show calculation using the appropriate mol:mol ratio from the above dissociation reaction. B) The solvent, water itself undergoes autodissociation/autoproteolysis and the equilibrium constant for that process (shown below), Kw, is 10-14 at 25 °C. Based on the dissociation reaction shown below and the Kw value provided, what is the concentration of hydronium coming ONLY from the solvent, water? Show calculation. C) Compare the hydronium concentrations calculated in questions 1A and 1B and identify which one is bigger and why? D) Calculate the pH of a solution containing 2X10-1 M sulfuric acid in water considering the hydronium generated from the acid as well as the solvent. Where pH = -log[H3O+]. E) Calculate the pH of the solution containing 2X10-1 M sulfuric acid in water considering the hydronium generated ONLY from the acid. Where pH = -log[H3O+]. F) Most common pH meters can only read two decimal places. Based on this information, would you, for all practical purposes, consider the hydronium concentration coming from water when considering the pH of a strong acid? Explain.
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To learn how to calculate ion concentrations in an aqueous solution of a strong diprotic acid. Sulfuric acid, H2SO4, is a strong acid. Its complete dissociation in aqueous solution is represented as: H2SO4 → H+ + HSO4- An HSO4- anion can dissociate further by: HSO4- ⇌ H+ + SO42- But the extent of dissociation is considerably less than 100%. The equilibrium constant for the second dissociation step is expressed as: Ka2 = [H+][SO42-]/[HSO4-] = 0.012 Part A: Calculate the concentration of H+ ions in a 0.010 mol L-1 aqueous solution of sulfuric acid. Express your answer to three decimal places and include the appropriate units. [H+] =

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Transcript

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00:01 Hello guys, let's get started with the solution.
00:06 In the question we are told that aspirin, which have a molecular formula of c9h8o4 is a monoprotein guide and 2 gram of it is dissolved in 0 .600 liter of the solution.
00:33 From the molecular formula, we can find the molar mass of aspirin which is 180 gram per mole.
00:51 So using this we can calculate the concentration of c9 h .8 .o .4, that is the aspirin or salicylic acid.
01:04 Concentration would be mass over molar mass times volume, which would give the value around 0 .0185 now the equation of dissociation for the following acid can be represented as c9 h8o4 plus water equilibrium h3o plus c9 h7o4 minus.
01:46 We can make use of the ica approach now which would be in terms of concentration unit would be molarity molar.
01:56 Initial concentration is 0 .0185 molar.
02:01 For products it would be 0.
02:03 Let's assume that the change in concentration is x.
02:10 So at equilibrium 185 minus x, here it would be x and x.
02:17 Now the k a can be written as x square upon 0 .0185 minus x.
02:32 So we are talking to told that the ph of the solution was found to have value 2 .6.
02:42 So here we can find the concentration of hydronium ion.
02:50 A concentration of hydronium ion would be equal to x which would be minus 2 .60 on solving.
03:00 It would come around 0 .00251 molar.
03:06 So, using the value of x, we can calculate the value of ka.
03:11 So, ka would be 0 .00251 whole square ratio 0 .0185 minus 0 .00251.
03:25 Upon solving the numbers, the value would come out to be around 3 .9 into 10 this to power minus 4...
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