Chemistry question - Estimate pH...?

Question

Estimate pH at the equivalence point of the reaction between 0.100 M solutions of ethanoic acid (CH3COOH) and sodium hydroxide (NaOH).

The answer is: Greater than 7... about 9

I just need help on how I get the answer. I'd really really appreciate anybody who could help me out.

Answer 1

you are correct it's 9.

how to get it?
pH = pKw - pKa(CH3COOH)
since Kw = 10^-14, pKw=14 and Ka(CH3COOH) = 1.8x10^-5 (better we round this into 10^-5 then pKa=5)
pH = 14-5 = 9

Answer 2

ok, let's see what I can do.

equivalence pt is a stoichiometric point at which just enough titrant has been added to totally consume the analyte. As we know acetic acid is a weak acid. At its equivalence pt, all acetic acid is consumed by the base, so what we have in soln is a netural cation and a _basic_ anion, acetate. The reason being is acetate is the conjugate base of acetic acid, as acetic acid being a weak acid, acetate should be a weak base. That's why the soln at equivalence point is basic.

so how to calculate pH at that point, ez. so, say we have 0.100M acetic acid and 0.100M NaOH. We need equal amount of both reactant to neutralize each other, right? So, without calculating, the concentration of acetate at equivalence point is 0.050M [the mole of acetate equals the intal moles of acetic acid, but the volume of the soln has doubled]. So, now comes the math part. pKa value for acetic acid is 1.74 E-5. You will need to do a basic acid/base equilibrium problem to figure out the concentration of OH at equivalence pt. pH = 14 - pOH

Answer 3

CH3COOH + NaOH -> CH3COONa + H2O

At the equivalence point all of the acid has reacted with base.
Both are monovalent and have the same concentration thus you need to add equal volumes of both solutions.
**** How to calculate this in general:
valency(acid)* M(acid)* V(acid)= valency(base)* M(base)* V(base)

thus 1*0.1* V(acid)= 1*0.1* V(base)
thus V(acid)=V(base)
***************

Since the stoichiometry is 1:1 you will have the same amount of mole CH3COONa forming but since the volume is doubled the concentration will be halfed, thus you have 0.05 M CH3COONa

**** How to calculate this in general:
Vfinal= V(acid) + V(base) = 2V(acid)
mole acid =M(acid)*V(acid)
From the stoichiometry of the reaction
1mole acid gives 1 mole CH3COONa
thus M(acid)*V(acid) mole acid give M(acid)*V(acid) mole CH3COONa

and the initial concentration of CH3COOH is
C=mole CH3COONa / Vfinal = M(acid)*V(acid) / 2V(acid)= M(acid)/2
****************************************
Since CH3COO- is the conjugate base of a weak acid it will give the reaction:

.. .. .. .. .. CH3COO- + H2O <=> CH3COOH + OH-
Initial .. .. .. .. 0.05
React .. .. .. . x
Produce .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. x.. .. .. .. .. x
At Equilibrium. 0.05-x .. .. .. .. .. .. .. .. .. x.. .. .. .. .. x


***Note: in this type of reactions we don't consider water for the equilibrium equation because it's concentration doesn't change much (water is your solvent)

Kb= [CH3COOH][OH-] / [CH3COO-]= x^2/(0.05-x)

But Kb= Kw/Ka
For acetic acid Ka=1.8*10^-5

Therefore 10^-14/(1.8*10^-5)= x^2/(0.05-x) =>
5.556*10^-10 =x^2/(0.05-x)

You can solve the quadratic equation or make an approximation. Let's assume that x<<0.05 and thus 0.05-x is effectively 0.05.
Then 5.556*10^-10 =x^2/0.5 and x= 5.27*10^-6
x is 4 orders of magnitude smaller than 0.1 so our assumption is justified and correct.
so [OH-]= x =5.27*10^-6
thus pOH= -log[OH]= -log(5.27*10^-6)= 5.28
and pH=14-pOH= 14 - 5.28 = 8.72

I must admit I overlooked the doubling of the volume in my first answer...

Answer 4

8