Consider a #"0.075 M"# solution of ethylamine (#"C"_2"H"_5"NH"_2#, #K_b = 6.4xx10^(-4)#). Calculate the hydroxide ion concentration of this solution, and calculate the pH?
You are given some solution of
Well, look at the
#K_a = (K_(w))/(K_(a))->K_a = ((1*10^-14))/((6.4*10^-4))=1.56*10^-11#
Now, for weak acids and weak bases we can't just plug in whatever concentration they give us to find the pH because weak acids and weak bases do not
But because I don't like ICE tables, I will set up the equilibrium expression and explain it to you step-by-step.
Equilibrium expressions are written as products over reactants.
Now, before the reaction proceeds, we have just the reactant,
What is the
Note: The change should always match up with the stoichiometric values given by the balanced equation. It's a 1:1 mole ratio so we are good.
#K_b = ([cancel"0+"x][cancel"0+"x])/([0.075cancel"-x"])#
#K_b = ([x][x])/([0.075])#
#(6.4*10^-4) = ([x^2])/([0.075])#
#(6.4*10^-4)[0.075] = [x^2]#
#sqrt((6.4*10^-4)[0.075]) = sqrt(x^2)#
#x = 0.006928, "so "color(orange)([OH^-] = 0.006928#
Remember when I crossed out the
Since we figured out the [C] of
#"pOH" = -log[0.006928" M"]#
Now from the following,
we can figure out the
#pH+pOH = 14#
#pH = 14-2.15 ->11.85#
#color(orange)(pH = 11.85)#
As a quick-trick in working with weak bases (ammonia or ammonia derivatives in pure water)
and for weak acids (monoprotic and the 1st ionization step of diprotic acids) in pure water
NOTE: These are good for weak electrolytes in pure water. If system is a common ion type problem, the ICE table needs to be used.
Given: (Wk Acid Problem)