Gen Chem Acid/Base Equalibrium..

[whc235]

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Im going insane with these questions.. im getting every single one wrong lol.. I dont understand them at all... balls. Would some1 mind doing a few and explain them??

1. A 50.0-mL sample of 0.50M acetic acid, HC2H3O2, is titrated with a 0.150 M NaOH solution. Calculate the pH after 25.0 mL of the base have been added (Ka = 1.8 x 10-5).

Correct Answer:
3.99

2. Calculate the pH of a buffer that is 0.20 M in formic acid and 0.15 M in sodium formate (Ka = 1.8 x 10-4).

Correct Answer:
3.62

3. Ksp for AgI = 8.3 x 10-17, and Kf for Ag(CN)2- = 1.0 x 1021. Calculate the equilibrium constant for the following reaction: AgI(s) + 2CN-(aq)

Ag(CN)2-(aq) + I-(aq)


Correct Answer:
8.3 x 10
😕😕😕😕😕👎 Help is much appreciated!!

 

[drdolittle1]

Would something like this be on the DAT? I feel like I would need a calculator.... I hate titrations

 

[pbure9]

I don't think you'll see something like 1 or 3. Those calculations seem way to hard and long and I'm actually not too sure how to go about doing them. It'd be nice if anyone could solve them.

I can help you with number 2 tho.

HForm is formic acid. So you have HForm --> H+ + Form-

All you have to do is set up the ka equation = [H+][Form-]/[HForm]
You have the concentration for formate (form-) = .15 and formic acid (Hform) is .20.
So we have 1.8 x 10-4 = ([H+] x .15)/.20
Without a calculator I'd do, 15/20 = 3/4. 4/3 x 1.8E-4 is about 2.5x10-4. Thats the [H+]. Do -log[H+] and my approximation would be 3.4 for a pH. Checking with my calculator now, it says 3.6. Close enough hopefully on the DAT

 

[Philippines03j]

Alright I had time to do one of them as of now. I came up with the right answer so i believe i did it correctly hope the following explanation is correct.
2. Calculate the pH of a buffer that is 0.20 M in formic acid and 0.15 M in sodium formate (Ka = 1.8 x 10-4).

The final equation we will end up using in this problem is PH=Pka+log(A-/HA)
----Lets start with the Pka which is calculated by -log(Ka) or -log(1.8x10-4)
The logs can be estimated as 3.8 since we wont use a calculator.
----Then all we know that the formic acid is the proton donor so its concentration goes in for HA and the sodium formate concentration goes in for A-.
----The equation should now look like PH=3.8+Log(.15/.2) or 3.8+Log(.75)
----Log(.75) or Log(7.5 x10-1)= about -.15.
----Ph=3.8-.15----> 3.65

Hope this helps!

looks like pbure9 explained it also...

 

[Philippines03j]

Would something like this be on the DAT? I feel like I would need a calculator.... I hate titrations

I don't think you'll see something like 1 or 3. Those calculations seem way to hard and long and I'm actually not too sure how to go about doing them. It'd be nice if anyone could solve them.

I can help you with number 2 tho.

HForm is formic acid. So you have HForm --> H+ + Form-

All you have to do is set up the ka equation = [H+][Form-]/[HForm]
You have the concentration for formate (form-) = .15 and formic acid (Hform) is .20.
So we have 1.8 x 10-4 = ([H+] x .15)/.20
Without a calculator I'd do, 15/20 = 3/4. 4/3 x 1.8E-4 is about 2.5x10-4. Thats the [H+]. Do -log[H+] and my approximation would be 3.4 for a pH. Checking with my calculator now, it says 3.6. Close enough hopefully on the DAT

How are you doing your estimations of logs? Because the way you did it I would have guessed the estimation of 3.65....
All I do is one less than the power so 3 just like you do but then on the other part I usually do 9 minus the number given. I think most say to do 10-the number but I found 9 gets you closer to the answer most of the time. So 9-2.5= 6.5 and the answer would be 3.65. If that helps....

 

[hham]

Titration problems screwed me over for my chem2 final, but it's funny I remember how to do it!!! Problem #1:

You first need to recognize that it's Weak-Acid/Strong Base Titration.
so the equation will be:

CH3COOH + NaOH ---> CH3COO- + H2O
You can see that the acetic acid is weak acid and therefore produces the acetate ion (CH3COO-) that will undergo hydrolysis.

You can see that NaOH doesn't show up on the product side b/c it's a strong base and it has disassociated completely, so there's really no need to bother showing it in the equation.

So let's set up the ice chart. For titration problems, you can convert everything to mmol/mol.

50mL of .50M Acetic Acid: 25mmol
25mL of .150M NaOH: 3.75mmol

---------CH3COOH + NaOH -------> CH3COO- + H2O
Initial ---25mmol------3.75mmol-------0
Change---3.75mmol---3.75mmol-----+3.75mmol
--------------------------------------------------
Eqlbm---21.25mmol-----0-----------+3.75mmol

*The reason for adding the 3.75mmol of NaOH to the (CH3COO-) Acetate Ion is b/c it has gone through complete disassociation.
So with these number of mmol, you can use the Henderson-Hesselbach Equation.

pH = pKa + log (Conjugate Base/Weak Acid)
and plus in the numbers

pH = 4.74 + log (3.75mmol/21.25mmol)
pH = 3.99