Question

In: Chemistry

Dissociation Constant For the dissociation reaction of a weak acid in water, HA(aq)+H2O(l)?H3O+(aq)+A?(aq) the equilibrium constant...

Dissociation Constant

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 2.62?

Express your answer numerically using two significant figures.

Ka =

SubmitHintsMy AnswersGive UpReview Part

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.

Kb =

Expert Solution

a) 2.00 g C9H8O4 x (1 mole C9H8O4 / 180 g C9H8O4) = 0.0111 moles C9H8O4

0.0111 moles C9H8O4 / 0.600 L = 0.0185 M C9H8O4

Molarity . . . . .C9H8O4 + H2O <==> H3O+ + C9H7O4-
Initial . . . . . . . .0.0185 . . . . . . . . . . . .0 . . . . . .0
Change . . . . . . . .-x . . . . . . . . . . . . . .x . . . . . .x
At Equil. . . . .0.0185 - x . . . . . . . . . . .x . . . . . .x

Ka = [H3O+][C9H7O4-] / [C9H8O4] = x^2 / (0.0185 - x)

pH = 2.60; [H3O+] = 10^-pH = 10^-2.60 = 0.0025 = x

Ka = (0.0025)^2 / (0.0185 - 0.0025) = 3.9 x 10^-4

b) Molarity . . . . . .C2H5NH2 + H2O <==> C2H5NH3+ + OH-
Initial . . . . . . . . . .0.100 . . . . . . . . . . . . . . .0 . . . . . .0
Change . . . . . . . . . .-x . . . . . . . . . . . . . . . .x . . . . . .x
At Equil. . . . . .0.100 - x . . . . . . . . . . . . . . .x . . . . . .x

pH = 11.87
pOH = 14.00 - pH = 14.00 - 11.87 = 2.13
[OH-] = 10^-pOH = 10^-2.13 = 0.0074 = x

Kb = [C2H5NH3+][OH-] / [C2H5NH2] = x^2 / (0.100 - x) = (0.0074)^2 / (0.100 - 0.0074) = 5.9 x 10^-4

queen_honey_blossom answered 2 months ago

What is the equilibrium constant expression for the following reaction: HCN(aq) + H2O(l)↔CN−(aq) + H3O+(aq) Choose...

What is the equilibrium constant expression for the following reaction: HCN(aq) + H2O(l)↔CN−(aq) + H3O+(aq) Choose from the list below and enter the letters alphabetical order. (e.g. For an equilibrium constant of [H3O+]-1eq[HCN]eq enter AH.) A) [HCN]eq E) [HCN]-1eq I) [Ru(NH3)62+]6eq B) [H2O]eq F) [H2O]-1eq J) [Ru2+]-6eq C) [CN-]eq G) [CN-]-1eq K) [NH3]-6eq D) [H3O+]eq H) [H3O+]-1eq L) [Ru(NH3)62+]-6eq

1. 3. Consider the dissociation of acetic acid: CH3COOH(aq) + H2O(l) ⎯→ ←⎯ CH3COO-(aq) + H3O+(aq)...

1. 3. Consider the dissociation of acetic acid: CH3COOH(aq) + H2O(l) ⎯→ ←⎯ CH3COO-(aq) + H3O+(aq) Ka = 1.8 x 10-5 pKa = 4.74 a. Write the equilibrium expression. b. Would the addition of an aqueous solution of sodium hydroxide favor the dissociation of acetic acid? Explain your answer. c. Would the addition of an aqueous solution of sodium acetate favor the dissociation of acetic acid? Explain your answer.

Hydrocyanic acid (HCN) is a weak acid that dissociates in water as follows. HCN(aq) + H2O(l)...

Hydrocyanic acid (HCN) is a weak acid that dissociates in water as follows. HCN(aq) + H2O(l) equilibrium reaction arrow H3O+(aq) + CN −(aq) The acid-dissociation constant (Ka) for hydrocyanic acid is 6.2 ✕ 10−10. Calculate the percent ionization (percent dissociation) if the initial concentration of hydrocyanic acid was 0.29 M.

Calculate the equilibrium constant for the following reaction of a weak acid and base HClO(aq) +...

Calculate the equilibrium constant for the following reaction of a weak acid and base HClO(aq) + NH3(aq) NH4+(aq) + ClO-(aq) given Ka (HClO) = 2.8 × 10-8, Kb (NH3) = 1.8 × 10-5 and Kw = 1.0 × 10-14.

In the titration of a weak acid, both a neutralization reaction and an acid‐dissociation equilibrium reaction...

In the titration of a weak acid, both a neutralization reaction and an acid‐dissociation equilibrium reaction with water (Ka) are occurring. Provide a molecular‐level explanation that describes how these two reactions contribute to the resulting pH observed during the drop‐by‐drop addition of NaOH during a titration procedure.

Write the equilibrium constant expression for the dissolution of ammonia in water. NH4+(aq) + H2O(l) <--->...

Write the equilibrium constant expression for the dissolution of ammonia in water. NH4+(aq) + H2O(l) <---> NH4+(aq) + OH- (aq) (In this reaction assume that the concentration of water will remain constant)

The weak acid HA with a dissociation constant Ka is distributed between and organic solvent and...

The weak acid HA with a dissociation constant Ka is distributed between and organic solvent and water. If the only extractable species is the undissociated species HA with a partition coefficient K, derive an expression showing the dependence of the distribution ratio, D, on [H+] of the aqueous phase and draw conclusions from this expression.

A weak acid, (HA), has an acid dissociation constant of 4.60∙10–6. A 25.00 ml sample with...

A weak acid, (HA), has an acid dissociation constant of 4.60∙10–6. A 25.00 ml sample with a concentration of 0.1200 M is titrated with 0.1500 M NaOH. F. How many ml NaOH have been added at the equivalence point? g. What is the pH at the equivalence point?

What is the equilibrium constant expression for the following reaction: C5H5NH(aq) + H2O(l)↔C5H5NH2+(aq)+OH−(aq) Choose from the...

What is the equilibrium constant expression for the following reaction: C5H5NH(aq) + H2O(l)↔C5H5NH2+(aq)+OH−(aq) Choose from the list below and enter the letters alphabetical order. (e.g. For an equilibrium constant of [OH-]-1eq[H2O]eq enter AH.) A) [H2O]eq E) [H2O]-1eq I) (XPbI2)2eq B) [C5H5NH]eq F) [C5H5NH]-1eq J) [Pb2+]-2eq C) [C5H5NH2+]eq G) [C5H5NH2+]-1eq K) [I-]-2eq D) [OH-]eq H) [OH-]-1eq L) (XPbI2)-2eq

The acid-dissociation constant for benzoic acid (C6H5COOH) is 6.3×10−5. A. Calculate the equilibrium concentration of H3O+...

The acid-dissociation constant for benzoic acid (C6H5COOH) is 6.3×10−5. A. Calculate the equilibrium concentration of H3O+ in the solution if the initial concentration of C6H5COOH is 5.1×10−2M B. Calculate the equilibrium concentration of C6H5COO− in the solution if the initial concentration of C6H5COOH is 5.1×10−2 M . C. Calculate the equilibrium concentration of C6H5COOH in the solution if the initial concentration of C6H5COOH is 5.1×10−2 M .