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?

2 Answers
May 16, 2017

Answer:

#[OH^-] = 0.006928" M"#

#"pH" = 11.85#

Explanation:

You are given some solution of #color(blue)"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)# provided. The #K_b# is called the base dissociation constant. It tells you to what extent the base, #"ethylamine"# will ionize in solution. Now, compare it with the #color(blue)(K_a# of its conjugate-acid pair, #color(blue)"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")#

  • #K_a = (K_(w))/(K_(a))->K_a = ((1*10^-14))/((6.4*10^-4))=1.56*10^-11#

#color(white)(----)#that #K_b>K_a#, so the solution will be basic

#---------------------#

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%# ionize in solution. That is why whenever we figure out the #pH# 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.

#---------------------#

#color(magenta)"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^-#

#color(magenta)"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])#

Now, before the reaction proceeds, we have just the reactant, #"ethylamine"#, and no products, meaning no #"ethylammonium"# or #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)])#

What is the #0# and #0.075#? Initially, we were given the concentration of #"ethylamine"# as #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 #0# 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")#

  • #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 #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#.

  • #"pOH" = -log[0.006928" M"]#

  • #"pOH"= 2.15#

Now from the following,

#color(white)(aaaaaaaa)"pH+pOH = 14" color(white)(aaa)"at "25^@C#

we can figure out the #"pH"#

  • #pH+pOH = 14#

  • #pH = 14-2.15 ->11.85#

  • #color(orange)(pH = 11.85)#

May 16, 2017

Answer:

#[OH^-] = 6.93 x 10^(-3)M# => #pOH = 2.16 => pH = 11.84#

Explanation:

As a quick-trick in working with weak bases (ammonia or ammonia derivatives in pure water)
=> #[OH^-] = sqrt((Kb)[Wk Base]# and pOH = -log[OH] & pH = 14 - pOH
and for weak acids (monoprotic and the 1st ionization step of diprotic acids) in pure water
=> #[H^+] = sqrt((Ka)[WkAcid]# => pH = -log[H]

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:
#[WkBase] = 0.075M ... &... K_b = 6.4 x 10^(-4)#

#[OH^-] = sqrt((0.075)(6.4x10^(-4))M# = #6.93 x 10^(-3)M#
#pOH = -log[OH^_] = -log(6.93 x 10^(-3))# = #2.16#
#pH = 14 - pOH = 14 - 2.16 = 11.84#

Given: (Wk Acid Problem)
#HA rightleftharpoons H^+ + A^-#
#[WkAcid] = 0.075M#; #K_a = 1.8 x 10^(-5)#
#[H^+] = sqrt((0.075)(1.8x10^(-5))#M = #1.16 x 10^(-3)M#
#pH = -log[H^+] = -log(1.16x10^(-3)) = 2.93#