# Chemistry

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A 40.00 mL sample of 0.1 M NH3 is titrated with 0.150 M HCl solution. Kb = 1.8 * 10^-5
What volume of base is required to reach the equivalence point?
What is the ammonium ion concentration at the equivalence point?
What is the pH of the solution at the equivalence point?

• Chemistry -

How much of this do you know how to do and exactly what is it you don't understand.?

• Chemistry -

I think that the first question you just do MaVa = MbVb (a is acid b is base). I have no idea how to do the second question, I'm guessing that it's just .1 M. I also know that pH = -log [H+], but do I use .15M? I'm just confused with the equivalence point.

• Chemistry -

right. That's how you do a.
Thre second part. You know how many mols it takes to neutralize the ammonia. That will be the mols of NH4Cl produced. And that divided by the volume (in liters which will be the sum of volume HCl + volume NH3) will be the molarity of the ammonium ion concn at the equivalence point (which is the neutralization point).
For the last part, you know the concn of the ammonium ion. Just hydrolye that (react with water) and do the ICE thing.
NH4^+ + HOH ==> NH3 + H3O^+

Ka = Kw/Kb = (H3O^+)(NH3)/(NH4^+)

• Chemistry -

What is the ICE thing? Is it pH = Ka + log [HA]/[A-]?

• Chemistry -

The ICE thing is difficult to show on these boards because we have a problem with spacing. But I think I can show you.
I re-read the problem and I think it means something different. I think it should have said, "what volume of ACID (not base) is required to reach the equivalence point?"
For a) you should have 26.67 mL HCl required to neutralize 40 mL of 0.1 M NH3 solution.
b)The ammonium ion concn at the equivalence point is 0.004 mols NH4^+/0.0667 L = 0.06 M

c)The NH4^+ hydrolyzes in water.
NH4^+ + H2O ==> NH3 + H3O^+
For the ICE we will have three rows
I
C
E
but then I can't space to put what these values are under the appropriate symbols. So what I will do is write out the formula and the concentration like this.
Initial concn (before rxn with H2O):
(NH4^+) = 0.06 M
(H3O^+) = 0
(NH3) = 0

Change in concn:
(H3O^+) = x
(NH3) = x
(NH4^+) = - x

Equilibrium:
(H3O^+) = 0 + x = x
(NH3) = 0 + x = x
(NH4^+) = 0.06 - x = 0.06 - x

Then Ka = Kw/Kb = (H3O^+)(NH3)/(NH4^+)

You know Kw and Kb. Substitute x for (H3O^+) and for (NH3) and substitute 0.06 - x for (NH4^+). solve for x which equals (H3O^+), then pH = - log (H3O^+),

You need not be intimidated with the word equivalence point. That is just the point at which the acid and base have exactly neutralized each other. In this case we have the reaction:
NH3 + HCl ==> NH4Cl
So we have no NH3 remaining, we have no HCl remaining (because both have been neutralized) and we have a solution of NH4Cl in water.

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