[specialk08]

Having a little trouble with all the numbers here..I don't think determining pH is my thing.....gets me so confused!!

1. 3.30g of (NH4)2SO4 is dissolved in 125.0mL of 0.1011M NaOH and diluted to 500.0mL
a) Find the pH of the solution.
b) Determine the new pH after 10.00mL of 0.2000M HCl is added to this solution.


Okay, so for this one I've determined I need to find molarity for the (NH4)2SO4 solution. So I calculate molarity by the # moles of solute (just change the 3.30g to moles) divided by the liters of solution. Would I use the 0.0125L as this value?

Anyway, once I find molarity for (NH4)2SO4, do I use the Henderson-Hasselbalch to calculate the pH?

So would it be pH = pKa of (NH4)2SO4 + log [(NH4)2SO4] ?
                                                            [NaOH]

I'm not sure if I am on the right track with that...

And for Part B I'm not sure where to begin....

2. 400.00mL of a TRIS buffer of pH 8.4 is required.
a) Calculate the required molar ratio of base to salt in the buffer.
b) How many GRAMS of the chloride salt of TRIS (FM = 157.6) should be added to 400.0mL of a 0.1200M solution of TRIS to get the desired pH?

- For convenience, write the formula of TRIS as "B" and that of the TRIS sald as "BHCl". The acid part of the salt is "BH+"


Okay so Question number 2 I am stumped on....

[Borek]

Obviously your main problem is in determining what to put into HH equation. Do you know what are conjugate acid and base?

[specialk08]

Yeah I'm just confused because the ones we did in class the equation had the acid/base and conj acid/conj base in it. :/

[Borek]

Try to guess then. What is ammonia? Does it have a conjugate acid or base? If so - what is it?

Hint in TRIS question should give you an instant answer, once you will try to write a general reaction equation involving conjugate acid, base and proton.

[specialk08]

Try to guess then. What is ammonia? Does it have a conjugate acid or base? If so - what is it?

Hint in TRIS question should give you an instant answer, once you will try to write a general reaction equation involving conjugate acid, base and proton.

Ok, so ammonia is a weak base, and has a conjugate acid of NH4+, right? I'm still confused with the SO4 that's present though.

And for part 2, would the equation look like this?

B + BHCl  B- + BH+ ?
I'm so confused

[Borek]

Ok, so ammonia is a weak base, and has a conjugate acid of NH4+, right? I'm still confused with the SO4 that's present though.

Good. SO₄^2- is just a spectator (not the only one in the solution).

And for part 2, would the equation look like this?

B + BHCl  B- + BH+ ?

No. Hydrochloride is in a way similar to ammonium chloride. Hydrochlorides in solutions are almost always 100% dissociated to chloride anion (spectator in this case) and R-NH₃⁺ - protonated amine (assuming it was amine hydrochloride, TRIS is an amine). So the reaction you are looking for will be the one involving R-NH₃⁺, written as BH⁺ - where B stands for R-NH₂.

[specialk08]

Ok, so ammonia is a weak base, and has a conjugate acid of NH4+, right? I'm still confused with the SO4 that's present though.

Good. SO₄^2- is just a spectator (not the only one in the solution).

And for part 2, would the equation look like this?

B + BHCl   B- + BH+ ?

No. Hydrochloride is in a way similar to ammonium chloride. Hydrochlorides in solutions are almost always 100% dissociated to chloride anion (spectator in this case) and R-NH₃⁺ - protonated amine (assuming it was amine hydrochloride, TRIS is an amine). So the reaction you are looking for will be the one involving R-NH₃⁺, written as BH⁺ - where B stands for R-NH₂.

Okay so for part 1 would the equation be this?


pH = pKa of (NH4+) + log (NH3)
                                   NH4+                                                


And I'm sorry, I am still quite confused on what you said about part 2.

[Borek]

pH = pKa of (NH4+) + log (NH3)
                                   NH4+                                                

Correct.

And I'm sorry, I am still quite confused on what you said about part 2.

for ammonia:

NH₃ + H⁺ <-> NH₄⁺

which can be written as

B + H⁺ <-> BH⁺

For TRIS solution B = TRIS, hydrochloride - when dissolved - gives BH⁺ and Cl^-. You are going to mix TRIS and TRIS hydrochloride - one of them is B other BH⁺, or one of them is a base, the other conjugate acid.

[specialk08]

pH = pKa of (NH4+) + log (NH3)
                                   NH4+                                                

Correct.

And I'm sorry, I am still quite confused on what you said about part 2.

for ammonia:

NH₃ + H⁺ <-> NH₄⁺

which can be written as

B + H⁺ <-> BH⁺

For TRIS solution B = TRIS, hydrochloride - when dissolved - gives BH⁺ and Cl^-. You are going to mix TRIS and TRIS hydrochloride - one of them is B other BH⁺, or one of them is a base, the other conjugate acid.

Ok, so for part one, if that is the correct set up of the equation..how do I know [NH3]?

[Borek]

NH₄⁺ + OH^- -> ...

[specialk08]

NH₄⁺ + OH^- -> ...

H2O + NH3+, but how does that give me a number?

[Borek]

NH₃, not NH₃⁺, but thats probably just a typo.

Simple stoichiometry - assume reaction went to the end.