CHE-1102 Lecture Notes - Lecture 6: Nitrous Acid, Ph, Ion

38 views2 pages

rubyporcupine166

24 Oct 2017

School

Department

Course

Professor

For unlimited access to Class Notes, a Class+ subscription is required.

Document Summary

Basic salt solutions: a salt derived from a strong base and a weak acid will form a basic solution, naoh + hno2 h2o + nano2. Na+ + no2: for any basic salt solution : a- + h2o ha + oh- 100: ex, determine the ph of a 2. 04 m solution of sodium acetate, determine the ions and their contribution to ph. C2h3o2- acetate (basic anion: write the equilibrium expression for a basic salt solution. A- + h2o ha + oh: make an equilibrium table for this solution, determine the value of kb. For salt solutions, assume you can always use the short cut! Determine the ph of the solution. poh = -log[oh-] = -log(3. 4x10-5) = 4. 47 ph = 14. 00 poh = 14. 00 4. 47 = 9. 53: dealing with basic salt solutions with ph > 7. 10: determine the ph an %hydrolysis of the solution made by dissolving 1. 00g of kf in 375ml of water, determine ions and their contribution to ph.

Related Questions

Salts formed by reaction of a strong acid with a weak base areacidic, and salts formed by reaction of a weak acid with a strongbase are basic. This is due to the equilibrium the conjugate haswith its acid or base. For example, sodium bicaronate NaHCO3contains the conjugate of the weak acid H2CO3. The conjugate willreform the acid through the equilibrium

HCO3?(aq)+H2O(aq)?H2CO3(aq)+HO?(aq)

and, thus, acts as a weak base. A solution of NaHCO3yields abasic solution.

If a salt is derived from a weak acid and a weak base, the pH ofthe solution depends on the relative acid strength (Ka) ofthe cation and base strength (Kb) of the anion. IfKa>Kb, the solution of the salt is acidic. IfKa<Kb, it is basic. If Ka?Kb, it is neutral.

Given that

Ka for HCN is 4.9

Calculate the pH of a solution containing a salt AcNa derived from a strong base (NaOH) and a weak acid (AcH), like CH₃COONa, KF, NaNO₂ and so on. In this case two processes have to be considered:

1. Dissociation of the salt, within the assumption that the salt is a strong electrolyte:

AcNa â Ac^- + Na⁺, for instance: CH₃COONa â CH₃COO^- + Na⁺

2. the hydrolysis of the water

Ac^- + H₂O â AcH + OH^-, for example: CH₃COO^- + H₂O â CH₃COOH + OH^-

knowing that the initial concentration of the salt is 0.17 M and the acid dissociation constant, K_a , is 1.08e-7.

pH=?

Analyze the pH of each salt solution: NaC2H3O2, NH4I, KHCrO4.

For KHCrO4 Ka1=1.5X10^-4, Ka2=1.0X10^-9

How do I do this without the Ka values for NaC2H3O2 and NH4I?

Example:

CHE-1102 Lecture Notes - Lecture 6: Nitrous Acid, Ph, Ion

Document Summary

Get access

Related Documents

CHE-1102 Lecture Notes - Lecture 7: Solubility Equilibrium, Titration Curve, Equivalence Point

CHE-1102 Lecture Notes - Lecture 7: Gastric Acid, Equilibrium Constant, Chief Operating Officer

CHE-1102 Lecture Notes - Lecture 9: Ph, Lewis Acids And Bases, Conjugate Acid

Related Questions