# Acid/bases equilibrium det. pH question

#### warcarrot

I just can't get this question right, I feel I'm missing some piece of understanding. Help me plz

H20(l)+CH3CO2Na (aq) = CH3CO2H (aq)+NaOH (aq)
What would be the pH of the solution if just enough HCl were added to the solution in the above reaction to consume all of the sodium acetate? and initial concentration of sodium acetate is 0.1 M. pKa of acetic acid is 4.74

I wrote this equation, I think it's right: 2CH3CO2Na+HCl+H20 = 2CH3CO2H + NaCl+ NaOH
and made an ICE table.. but I keep getting a very basic pH, but the answer says that the resulting pH is 2.87, I just don't understand how. Can someone walk through it? I'd really appreciate it, thanks <3

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#### jetsfan1234

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Think of it this way: when you add hcl, water is no longer the strongest acid, so it ceases to react.Essentially you end up with h2o and ch3co2h. No naoh is formed. Final ph is dependent solely on the concentration of ch3co2h.

OP
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#### warcarrot

Ok so I now have CH3CO2Na + HCl = CH3CO3H + NaCl
Do I know the initial concentration of HCl?
because all the sodium acetate is reacted away, is the final equilibrium 0 M so the change is 0.1?
I'm confused on what I know/don't know....to find the final eq concentration of CH3CO3H..

#### jetsfan1234

##### Membership Revoked
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Ok so I now have CH3CO2Na + HCl = CH3CO3H + NaCl
Do I know the initial concentration of HCl?
because all the sodium acetate is reacted away, is the final equilibrium 0 M so the change is 0.1?
I'm confused on what I know/don't know....to find the final eq concentration of CH3CO3H..
How much sodium acetate do you have to begin with? You assume all of it reacts with the hcl. So moles of sodium acetate at start = moles of acetic acid after addition of hcl.

When you calculate final molarity of acetic acid, make sure you account for the volume of hcl that was added to solution. Edit-this might be insignificant, depending on the problem.

Then you use the Ka of acetic acid to find the final ph.

Equation: ch3co2h ---> ch3co2- + h+ (this is where the ICE chart can be used)

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#### Lifeman

10+ Year Member
Whenever you have JUST a weak acid or weak base added to water you use an ICE chart. If you have a mixture of conjugates, you use H-H equation. So in your problem, to find the pH when it was just acetate you would have used the ICE chart. When you started adding HCl you would have to use H-H, but when you convert all of the weak base into weak acid using HCl, you just have a weak acid in solution so its back to using ICE. It's going to be 2 x 10^-5 = x^2 / .1

#### Odi

Initial concentration of sodium acetate is 0.1 M. When they add HCl, you'll end up with 0.1 M acetic acid. That should make intuitive sense as 0.1 M of H+ ions are consumed by your starting concentration of 0.1 M sodium acetate.

The problem becomes simple... you have 0.1 M of a weak acid with a pKa of 4.74... find the pH?

1) Find the Ka first --> pKa=-log[Ka] --> 4.74=-log[Ka] --> thus Ka=10^-4.74

2) Set up your equilibrium expression:

Ka=[H+][CH3COO-]/[CH3COOH]

The denominator is [0.1M - x] ...because you start off with 0.1 M CH3COOH and it dissociates (thus the "- x"). But since it's a weak acid the amount it dissociates is negligible to the original concentration so you drop the "-x." And you should know that [H+] and [CH3COO-] will equal each other when the weak acid dissociates so they are [x][x]=x^2.

Thus your equilibrium expression is: 10^-4.74=x^2/0.1

Solve for x which is your concentration of H+ ions. Then calculate the pH from there with pH=-log[H+].

• Czarcasm
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#### warcarrot

CH3CO2- + H+ = CH3CO2H
I 0.1 x 0
C -0.1 -0.1 +0.1
E 0 x-0.1 0.1(HCl?)

the problem didn't give the actual vol of hcl added, but I know it is significant, because the answer is pH 2.74, how do I incorporate the HCl vol?

Edit: okay wait what i just said might be irrelevant let me figure out these other comments first....

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#### jetsfan1234

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CH3CO2- + H+ = CH3CO2H
I 0.1 x 0
C -0.1 -0.1 +0.1
E 0 x-0.1 0.1(HCl?)

the problem didn't give the actual vol of hcl added, but I know it is significant, because the answer is pH 2.74, how do I incorporate the HCl vol?
First off, be careful with the ice chart. You start with 0.1 mole of acid, but that doesn't all dissociate completely, because organic acids are weak acids. You are trying to determine how much acetic acid dissociates:

Ch3co2h ---> h+ + ch3co2-

I 0.1. 0. 0

C - X X. X

E 1-X X. X.

Then use Ka = (Xsquared)/(1-X). x gives you h+ concentration, and from there you calculate ph

If you need to incorporate vol of hcl: did they give you molarity of hcl solution? You need 0.1 moles hcl to react. You can figure out volume needed from the molarity.

OP
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#### warcarrot

Initial concentration of sodium acetate is 0.1 M. When they add HCl, you'll end up with 0.1 M acetic acid. That should make intuitive sense as 0.1 M of H+ ions are consumed by your starting concentration of 0.1 M sodium acetate.

The problem becomes simple... you have 0.1 M of a weak acid with a pKa of 4.74... find the pH?

1) Find the Ka first --> pKa=-log[Ka] --> 4.74=-log[Ka] --> thus Ka=10^-4.74

2) Set up your equilibrium expression:

Ka=[H+][CH3COO-]/[CH3COOH]

The denominator is [0.1M - x] ...because you start off with 0.1 M CH3COOH and it dissociates (thus the "- x"). But since it's a weak acid the amount it dissociates is negligible to the original concentration so you drop the "-x." And you should know that [H+] and [CH3COO-] will equal each other when the weak acid dissociates so they are [x][x]=x^2.

Thus your equilibrium expression is: 10^-4.74=x^2/0.1

Solve for x which is your concentration of H+ ions. Then calculate the pH from there with pH=-log[H+].
Ah it makes more sense now.. writing it as a dissociation reaction makes it easier..
Thank you everybody!

• Odi

#### Czarcasm

##### Hakuna matata, no worries.
5+ Year Member
The denominator is [0.1M - x] ...because you start off with 0.1 M CH3COOH and it dissociates (thus the "- x"). But since it's a weak acid the amount it dissociates is negligible to the original concentration so you drop the "-x." And you should know that [H+] and [CH3COO-] will equal each other when the weak acid dissociates so they are [x][x]=x^2.
@jetsfan1234 @Odi It's been a while since I've looked at this, and for the most part I get it -- but I'm a little confused as to why we would consider the "x" term in (0.1-x) as negligible. Wouldn't we expect 100% of the starting weak base to react completely with the strong acid. That is "x" is 0.1 M, and therefore the overall the final concentration of weak base is zero (since it dissociates completely in the presence of strong acid)?

Maybe my understanding of what x represents is a bit flawed.

#### Czarcasm

##### Hakuna matata, no worries.
5+ Year Member
EDIT: This is essentially what they said above -- I just didn't understand it until after I typed this up lol.
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I would of approached this a little differently. Knowing HCl would react with the base completely to produce the conugate acid, I'd instead focus on the equilibrium of the conjugate acid.

So we know that after reacting the base with strong acid, we have: 0.1 M CH3CO2H and all the conjugate base we had initially was consumed. The only conjugate base present after the fact, would be produced from the equilibrium reaction of this new weak acid in our solution.

Now, considering the equilibrium expression of this reaction, we have:

CH3CO2H + H2O(l) ---> CH3CO2- + H3O+

Using ICE Chart to find equilibrium concentrations:

CH3CO2H(aq) + H2O(l) ---> CH3CO2-(aq) + H3O+(aq)

I: 0.1 -- 0 + 0
C: -x -- +x +x
E: 0.1-x -- +x +x

The Ka expression would then be: [CH3CO2-][H3O+]/[CH3CO2H] (Ignoring liquids/solids)

Substituting the values from ICE chart:

Ka = [x][x]/[0.1] (I think this is where we assume -x is negligble since by definition, a weak acid will barely dissociate -- compared to its initial concentration).

And then we simply solve for x using the Ka value. And then finally, because x = concentration of H30+, we take the -log of that concentration to find pH and viola. Yay, I get it!

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• Odi

#### jetsfan1234

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@jetsfan1234 @Odi It's been a while since I've looked at this, and for the most part I get it -- but I'm a little confused as to why we would consider the "x" term in (0.1-x) as negligible. Wouldn't we expect 100% of the starting weak base to react completely with the strong acid. That is "x" is 0.1 M, and therefore the overall the final concentration of weak base is zero (since it dissociates completely in the presence of strong acid)?

Maybe my understanding of what x represents is a bit flawed.
There are 2 essential steps to the problem:
1. Hcl reacts completely with sodium acetate to form acetic acid. (so, 0.1 moles acetate---> 0.1 moles acetic acid)
2. Acetic acid reaches equilibrium. It is the equilibrium of the weak acid (acetic acid) that determines the final ph of the solution. THIS is the step that requires the ice chart.

Weak bases do not dissociate very much --only strong acids (eg, hcl) can be assumed to dissociate completely. For example, If you have 0.1 moles of a weak acid, maybe .0002 moles will dissociate. So the x in the 0.1-x term Is negligible from a mathematical perspective. It makes calculation of [h+] a lot easier (no quadratic equation required).

• Czarcasm

#### Czarcasm

##### Hakuna matata, no worries.
5+ Year Member
There are 2 essential steps to the problem:
1. Hcl reacts completely with sodium acetate to form acetic acid. (so, 0.1 moles acetate---> 0.1 moles acetic acid)
2. Acetic acid reaches equilibrium. It is the equilibrium of the weak acid (acetic acid) that determines the final ph of the solution. THIS is the step that requires the ice chart.

Weak bases do not dissociate very much --only strong acids (eg, hcl) can be assumed to dissociate completely. For example, If you have 0.1 moles of a weak acid, maybe .0002 moles will dissociate. So the x in the 0.1-x term Is negligible from a mathematical perspective. It makes calculation of [h+] a lot easier (no quadratic equation required).
Ahaha, I just got it when I tried reasoning it out. Your explanation helped clarify things a bit better, thanks again.

• jetsfan1234