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Pka acidity


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Gotta Love Equilibrium.

Because my mind is about to explode but I shouldn't be slacking off because my grade depends heavily on this exam. Let's talk equilibrium.

Important things to remember when dealing with equilibrium problems:
Large k = mostly product (product favored)Small k = mostly reactants (reactant favored)Q>K is proceeds towards reactantsQ Acids are H+ donorsBases are H+ acceptorsQuadratic Formula: x = -b +/- Square root of b^2 - 4ac divided by 2aFactors that affect equilibrium: 1) Change in [ ] of reactants/products 2) Change in presssure 3) Change in temperature. Catalysts do not affect equilibrium
Kw = Ka*KbpH and pOH have a difference of 14pH = -log [H3o+]pOH = -log [OH-]pKa = -logKapH = pKa + log(A-/HA)I've always had difficulty with predicting which way equilibrium reactions would shift. But now that I've really looked into and broken it down. It's pretty simple. Let's take a look at this sample problem.
Consider the following reaction and predict which way the reaction will shift:HNO2 + ClO4- HClO4 + NO2-

First, we have to figure out which ones are the acids and which ones are the bases. As we recall from our notes above, acids are H+ donors. Which of those things in the rxn donate an H+ proton? HNO2 does because ClO4- doesn't even have an H+ proton. So if HNO2 is an acid, then that means NO2- in the products is the conjugate base.

Now that means Clo4- is a base and HClO4 is the conjugate acid. Now we have our two acids which are:
HNO2 and HClO4
Acid and Conjugate Acid

Looking at the Ka value, because Ka values determine acidity which determine acid strength. The higher the Ka value, the stronger the acidity is. The Ka of HNO2 is 7.2x10^-4. HClO4 is Perchloric acid which is a very strong acid and has a huge Ka value. Thus, we confirm that on the reactants side, the acidity is weaker (HNO2) and the acidity on the products side is much stronger (Perchloric acid).

Since the acidity strength...

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