You are given some solution of color(blue)"ethylamine"ethylamine, an organic molecule, which gives off a basic solution in pure water. How do I know the solution will be basic and not acidic?
Well, look at the color(blue)(K_b)Kb provided. The K_bKb is called the base dissociation constant. It tells you to what extent the base, "ethylamine"ethylamine will ionize in solution. Now, compare it with the color(blue)(K_aKa of its conjugate-acid pair, color(blue)"ethylammonium"ethylammonium. We know from the following relationship
color(blue)(K_a * K_b = K_w,color(white)(aa) K_w = 1*10^-14 ("water dissociation constant")Ka⋅Kb=Kw,aaKw=1⋅10−14(water dissociation constant)
- K_a = (K_(w))/(K_(a))->K_a = ((1*10^-14))/((6.4*10^-4))=1.56*10^-11Ka=KwKa→Ka=(1⋅10−14)(6.4⋅10−4)=1.56⋅10−11
color(white)(----)−−−−that K_b>K_aKb>Ka, so the solution will be basic
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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 100%100% ionize in solution. That is why whenever we figure out the pHpH of a weak acid or base, we have to use something called an ICE table .
But because I don't like ICE tables, I will set up the equilibrium expression and explain it to you step-by-step.
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color(magenta)"Step 1: Write out the reaction for the ionization of ethylamine in solution"Step 1: Write out the reaction for the ionization of ethylamine in solution
color(white)(----)C_2H_5NH_2 + H_2O rightleftharpoons C_2H_5stackrel(+)NH_3 + OH^-−−−−C2H5NH2+H2O⇌C2H5+NH3+OH−
color(magenta)"Step 2: Set up an equilibrium expression"Step 2: Set up an equilibrium expression
Equilibrium expressions are written as products over reactants.
color(white)(----)K_b = ([C_2H_5stackrel(+)NH_3][OH^-])/([C_2H_5NH_2])−−−−Kb=[C2H5+NH3][OH−][C2H5NH2]
Now, before the reaction proceeds, we have just the reactant, "ethylamine"ethylamine, and no products, meaning no "ethylammonium"ethylammonium or OH^-OH− ions. As the reaction proceeds, however, the reactants will decrease and products will start forming. This is the change denoted by the color(red)(+|-)+∣− signs. The color(red)(+)+ indicates increasing in amount and the color(red)(-)− indicates decreasing in amount.
K_b = ([C_2H_5stackrel(+)NH_3][OH^-])/([C_2H_5NH_2])->K_b = ([0color(red)(+x)][0color(red)(+x)])/([0.075color(red)(-x)])Kb=[C2H5+NH3][OH−][C2H5NH2]→Kb=[0+x][0+x][0.075−x]
What is the 00 and 0.0750.075? Initially, we were given the concentration of "ethylamine"ethylamine as 0.075" M"0.075 M. This concentration indicates the amount you had before the reaction reached equilibrium. Notice also that the product concentrations were not initially given so the 00 indicates just that.
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.
color(magenta)("Step 3: Find the "OH^(-) "concentration")Step 3: Find the OH−concentration
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K_b = ([cancel"0+"x][cancel"0+"x])/([0.075cancel"-x"])
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K_b = ([x][x])/([0.075])
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(6.4*10^-4) = ([x^2])/([0.075])
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(6.4*10^-4)[0.075] = [x^2]
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sqrt((6.4*10^-4)[0.075]) = sqrt(x^2)
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x = 0.006928, "so "color(orange)([OH^-] = 0.006928
Remember when I crossed out the color(red)(-x) for the change in concentration for ethylamine? I made the assumption that the change is so small that it would not make any significant difference. This is the 5% rule. If our percent ionization is less than 5%, then the assumption is valid.
"x"/("initial "[C]")*100% ->(0.006928)/(0.075)*100%= 0.093%
color(white)(aaaaaaaaaaaaaaaaaaaa)0.093% < 5%
color(magenta)("Step 4: Find the pH")
Since we figured out the [C] of OH^- ions, we can go ahead and find the pOH.
Now from the following,
color(white)(aaaaaaaa)"pH+pOH = 14" color(white)(aaa)"at "25^@C
we can figure out the "pH"
