Arrhenius, Bronsted-Lowry, & Lewis Models of Acids and Bases and Solving Acid/Base Problems. Chapter 14 Spring 2010. Arrhenius Acids and Bases. Arrhenius acids form hydrogen ions in aqueous solution with Arrhenius bases forming hydroxide ions. Bronsted-Lowry Acids and Bases.
Arrhenius, Bronsted-Lowry, & Lewis Models of Acids and Basesand Solving Acid/Base Problems
How to Calculate pH of strong acids
pH = -log (0.01) = 2
The pH of Weak Acids
Ka = 3.5 x 10-8
The shape of ICl4 is square planar due to the bonds of the Iodine atom to 4 chlorine atoms.
The shape of BF4 is tetrahedral because of the presence of more valence electrons, which bend the shape and make it tetrahedral instead of square planar even though like ICl4, there are 4 bonds to the surrounding atoms.
Two key features of polyprotic acids are that they lose their protons in a stepwise manner and that each proton is characterized by a different pK a. The factors contributing to the pK a of each acidic proton in a polyprotic species are the same factors that determine the relative acidity of monoprotic acids--the dominant factor is strength of the acid-H bond. Consider, for example, the triprotic acid H3PO4 shown in :
As each proton is lost from phosphoric acid, the phosphorous becomes more electron rich, and less electron withdrawing. Therefore, the loss of each proton strengthens the O-H bond and increases the pK a of the phosphate species. This trend is evident in the pK a data given in . In general, it is true that K a1, K a2, K a3, and so on, for polyprotic acids.
As you may have guessed, calculating the pH of a polyprotic acid solution is not as simple as it is for monoprotic acids. In fact, it is quite a messy problem. However, that mess can be quickly cleaned up by making the assumption, as we did for a mixture of acids, that only the strongest acid (i.e. only the first dissociation) has a significant effect on the pH.
Ka = 6.2 x 10-10
OH- > CN- > H2O
Salt is conjugate of either an acid or a base (sometimes both) according to Bronsted- Lowry
NH4NO3 (s) → NH4+ (aq) + NO3– (aq)
NH4+ can donate an H+, so NH4+ is an acid and NH3 is its conjugate base.
NO3–can accept an H+, so NO3– is a base and HNO3 is its conjugate acid.
In water NH4+ would be NH4+ + H2O → NH3 + H3O+
This reaction occurs, since NH4+ is a weak acid and NH3 is a weak base
Result is to increase the H3O+ concentration, so the solution becomes acidic.
KBr (s) → K+ (aq) + Br – (aq)
K+ is not a conjugate acid or a base.
Br – can accept an H+, so Br – is a base and HBr is its conjugate acid.
Br – + H2O → HBr + OH–
But this reaction does not occur to any reasonable extent because HBr is a strong acid, so the equilibrium in the reaction lies very strongly to the left.
Since no new H+ or OH– are produced, the resulting KBr (aq) solution remains as neutral as the original water solvent.