Chapter 14 Acids and Bases. Some Definitions. Arrhenius Acid: Substance that, when dissolved in water, increases the concentration of hydrogen ions. Base: Substance that, when dissolved in water, increases the concentration of hydroxide ions. Some Definitions. Brønsted – Lowry
Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.
A Brønsted–Lowry acid…
…must have a removable (acidic) proton.
A Brønsted–Lowry base…
…must have a pair of nonbonding electrons.
...it is amphiprotic.
In any acid-base reaction, the equilibrium will favor the reaction that moves the proton to the stronger base.
HCl(aq) + H2O(l) H3O+(aq) + Cl−(aq)
H2O is a much stronger base than Cl−, so the equilibrium lies so far to the right K is not measured (K>>1).
C2H3O2(aq) + H2O(l)
H3O+(aq) + C2H3O2−(aq)Acid and Base Strength
Acetate is a stronger base than H2O, so the equilibrium favors the left side (K<1).
Using the following Ka values, arrange the following species according to their strength as bases:
H2O(l) + H2O(l)
H3O+(aq) + OH−(aq)Autoionization of Water
Kc = [H3O+] [OH−]
pH is defined as the negative base-10 logarithm of the hydronium ion concentration.
pH = −log [H3O+]
Kw = [H3O+] [OH−] = 1.0 10−14
[H3O+] = (1.0 10−14)1/2 = 1.0 10−7
pH = −log (1.0 10−7) = 7.00
These are the pH values for several common substances.
[H3O+] [OH−] = Kw = 1.0 10−14,
we know that
−log [H3O+] + −log [OH−] = −log Kw = 14.00
or, in other words,
pH + pOH = pKw = 14.00
pH = -log10(H3O+)
pOH = -log10(OH-)
Relationship between pH and pOH
pH + pOH = 14
Finding [H3O+], [OH-] from pH, pOH
[H3O+] = 10-pH
[OH-] = 10-pOH
For more accurate measurements, one uses a pH meter, which measures the voltage in the solution.
[H3O+] = [acid].
HA(aq) + H2O(l)
A−(aq) + H3O+(aq)Dissociation Constants
the equilibrium expression would be
The greater the value of Ka, the stronger the acid.
Ka =Calculating Ka from the pH
pH = −log [H3O+]
2.38 = −log [H3O+]
−2.38 = log [H3O+]
10−2.38 = 10log [H3O+] = [H3O+]
4.2 10−3 = [H3O+] = [HCOO−]
Now we can set up a table…
[4.2 10−3] [4.2 10−3]
Ka =Calculating Ka from pH
= 1.8 10−4
These salts simply dissociate in water:
KCl(s) K+(aq) + Cl-(aq)
The basic anion can accept a proton from water:
C2H3O2- + H2O HC2H3O2 + OH- base acid acid base
The acidic cation can act as a proton donor:
NH4+(aq) NH3(aq) + H+(aq)
Acid Conjugate Proton
AlCl3(s) + 6H2O Al(H2O)63+(aq) + Cl-(aq)
Salt water Complex ion anion
Al(H2O)63+(aq) Al(OH)(H2O)52+(aq) + H+(aq)
Acid Conjugate base Proton
Calculate the pH of a 0.30 M NaF solution. The Ka value for HF is 7.2 x 10-4
Calculate the pH of a 0.10 M NH4Cl solution. The Kb value for NH3 is 1.8 x 10-5
Calculate the pH of a 0.010 M AlCl3solution. The Kavalue for Al(H2O)63+ is 1.4 x 10-5
Predict whether an aqueous solution of each of the following salts will be acidic, basic, or neutral.