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
Answer:
Explanation:
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 stepbystep.
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"]# 
#"pOH"= 2.15#
Now from the following,
we can figure out the

#pH+pOH = 14# 
#pH = 142.15 >11.85# 
#color(orange)(pH = 11.85)#
Answer:
Explanation:
As a quicktrick 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:
Given: (Wk Acid Problem)