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13 Dec 2019

Acids donate protons to water according to the general equation: HA(aq)+H2O(l)⇌H3O+(aq)+A−(aq) Consider the following acids and their equilibrium constants for reaction with water at 25°C. If all the acids have the same initial concentration, which is the strongest acid (i.e. which donates the most protons to water)?

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Jamar Ferry
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17 Dec 2019

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Related questions

A weak acid dissociates in water according to this equation (also found in the procedure):

HA(aq) + H2O(l) <--> H3O+(aq) + A-(aq)

Based on what you know about weak acids, which one of the following would be a reasonable value for Ka?

A) .0050
B) 5,000
C) 500
D) 1.0

For the dissociation reaction of a weak acid in water,

HA(aq)+H2O(l)⇌H3O+(aq)+A−(aq)

the equilibrium constant is the acid-dissociation constant, Ka, and takes the form

Ka=[H3O+][A−][HA]

Weak bases accept a proton from water to give the conjugate acid and OH− ions:

B(aq)+H2O(l)⇌BH+(aq)+OH−(aq)

The equilibrium constant Kb is called the base-dissociation constant and can be found by the formula

Kb=[BH+][OH−][B]

When solving equilibrium-based expression, it is often helpful to keep track of changing concentrations is through what is often called an I.C.E table, where I. stands for Initial Concentration, C. stands for Change, and E. stands for Equilibrium Concentration. To create such a table, write the reaction across the top creating the columns, and the rows I.C.E on the left-hand side.

Initial (M)Change (M)Equilibrium (M)A+ B→AB

Part A

Aspirin (acetylsalicylic acid, C9H8O4) is a weak monoprotic acid. To determine its acid-dissociation constant, a student dissolved 2.00 g of aspirin in 0.600 L of water and measured the pH. What was the Ka value calculated by the student if the pH of the solution was 2.60?

Express your answer numerically using two significant figures.

Hints
Ka =

SubmitMy AnswersGive Up

Incorrect; Try Again; 3 attempts remaining

You used the initial concentration of aspirin in the Ka expression. Instead, you need to use the equilibrium concentration. Consider that the amount of aspirin that reacted is equal to the amount of H3O+ produced. How much is left over?

Part B

A 0.100 M solution of ethylamine (C2H5NH2) has a pH of 11.87. Calculate the Kb for ethylamine.

Express your answer numerically using two significant figures.

Hints
Kb =

What is the hydroxide concentration, [OH-], in a 3.85 M NaC2H3O2 solution? Consider the following information to help you answer this question: Sodium acetate is a water-soluble salt. Like all water-soluble ionic compounds, it dissociates completely when dissolved in water: H2O NaC2H3O2(s) ---------------> Na+(aq) + C2H3O2-(aq) Since the Na+ ion is the conjugate acid of a strong base (NaOH) it is too weak of an acid to react with water: Na+(aq) + H2O(l) ---------> N.R. Note: N.R. means “No Reaction”. However, the C2H3O2- ion is the conjugate base of a weak acid (HC2H3O2). Therefore, it is a strong enough base to react with water: C2H3O2-(aq) + H2O(l) <----------> HC2H3O2(aq) + OH-(aq) The above equilibrium reaction is the one for which you need to set up and solve an ICE table. However, the Kb value for C2H3O2- is normally not published in tables. It doesn’t need to be, because you can get it from two other values that are published: the Ka for HC2H3O2 and KW for the ionization of water. HC2H3O2(aq) + H2O(l) <----------> H3O+(aq) + C2H3O2-(aq) Ka = 1.8 x 10-5 2H2O(l) <----------> H3O+(aq) + OH-(aq) KW = 1.0 x 10-14 By combining the above two equations in the appropriate way, you can produce the reaction for which the equilibrium constant is desired. Once you have the required equilibrium constant, you can set up the ICE table in the usual way. Note that the initial concentration of C2H3O2- is the concentration given for NaC2H3O2. This is because the reaction proceeds to 100% completion, and the reaction has one-to-one stoichiometry. For every mole of NaC2H3O2, you get a mole of C2H3O2- (and a mole of Na+).

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