Tue 2/20

SALTS of weak acids and weak bases are made by neutralization reactions. Equal amounts (moles) of strong acid and weak base combine to make a salt of a weak base (eg. BHCl). Equal amounts of strong base and weak acid are combined to make a salt of a weak acid (eg. NaX).

A helpful formula for equating moles of acid and moles of base when doing volumetric analysis (the neutralization) is

\[M_1V_1=M_2V_2\]

\(M\) is molarity and \(V\) is volume. That formula works for dilutions where original concentration and volume are \(M_1V_1\) and the final concentration and volume is \(M_2V_2\). Now just change the 1's to "acid" and the 2's to "base" and you have the correct formula for adding the correct amounts of acid and base for a 100% neutralization.

\[M_{\rm acid}V_{\rm acid}=M_{\rm base}V_{base}\]

All Type 1 problems (one species is put into water) are solved with the same looking RICE table and the equation to solve is always:

\[K = {x^2\over C-x}\]

That leads to a quadratic equation to solve and you get \(x = {\sqrt{K^2+4KC}-K\over 2}\)

IF \({C\over K} > 1000\) , then the assumption can be made that \(C-x\approx C\) which leads to the far simpler \(x = \sqrt{KC}\)

Use the RIGHT K! You need a \(K_{\rm a}\) for all acids which are both plain weak acids (HA) and also conjugate acids of weak bases (BH⁺). You need a \(K_{\rm b}\) for all bases which are both plain weak bases (B) and also conjugate bases of weak acids (A^-). You have to calculate the \(K_{\rm a}\) and \(K_{\rm b}\) for the ionic conjugate pairs by using the following relation:

\[K_{\rm w} = K_{\rm a} K_{\rm b}\]

Thu 2/15

For all monoprotic strong acids, you can just set the concentration directly equal to the hydrogen ion concentration.†

For all Group-1 hydroxides, MOH (strong base), you can just set the concentration directly equal to the hydroxide ion concentration.†

For all Group-2 hydroxides, M(OH)₂ (strong base), you can double the concentration and set that directly equal to the hydroxide ion concentration. Note: you double the concentration because there are 2 moles of hydroxide in every 1 mole of the base.†

† In each of the above cases, you do have to make sure that the stated concentration is greater than the background H⁺ and OH^- concentrations of 10^-7 M (neutral water, pH 7).

Since for Exam 2 we only put one species (acid or a base) into solution and nothing else, all equilibria will have the same one species splits into two species equilibria and RICE table such that the algebra is always:

\[K = {x^2\over C-x}\]

and if \({C\over K} > 1000\) (the assumption), then the algebra reduces to:

\[K = {x^2\over C}\]

This equation easily solves with \(x = \sqrt{KC}\) where \(x\) = [H⁺] for acid problems and \(x\) = [OH^-] for base problems.