Posted by Ashley on Sunday, March 10, 2013 at 10:00pm.
Ashley, tel me what your problem is so I will know how to answer these.
But to get you stated, if you add an acid the soln will become more acid. If you add a base the soln will become more basic.
When you say "effect" is it enough to say it becomes more acid or more basic or are you to calculate the pH at the new conditions. If the latter, I can show you how to do one of them and the others follow.
i am suppose to calculate the pH at the new conditions, but i am super lost. also for c and d i think you would have to figure out the moles, but then i do not know what to do from there.
i am suppose to calculate the pH at the new conditions, but i am super lost. also for c and d i think you would have to figure out the moles, but then i do not know what to do from there.
Thanks.
You have 1L of a 1 M solution of NaHC2O4 and Na2C2O4 a a pH = 4.0
a. You add 1g NaHC2O4 to this.
First you must calculate what you have in the buffer. You do this as follows:
base(B) + acid(A) = 1M
That's two unknowns so you need a second equation. That one is
pH = pK2 (for H2C2O4) + log (B)/(A)
4.0 = 4.27 + log(B)/(A)
Look up pK2 for H2C2O4 and use that value from your notes/text and not the one above. Texts differ and mine is several years old.
There are your two equations in two unknowns. Solve for A and B.
Then add the 1 g (converted to mols and M of course) NaHC2O4
Here is what I get.
4.0 = 4.27 + log b/a
b/a = 0.513 pr b = 0.513a
Then
a + b = 1
a + 0.513a = 1
1.513a = 1
a = 0.661 M
b = 1-0.661 = 0.339 M
Therefore, the original solution consists of 1L of 0.661M NaHC2O4 and 0.339M Na2C2O4. You may obtain slightly different numbers if you use a different pKa2 than I used above. Check my work. The a part of the problem follows:
1g NaHC2O4 = 1/molar mass = approximately 0.00893 in 1L which makes 0.00893M.(It appears I added molarity; actually I multiplied 1L x M, added moles, and divided by 1L which is the same as adding molarities). We had 0.661M NaHC2O4. Total mols/L = 0.661+0.00893 = 0.670
(It may appear I added molarity; actually I multiplied 1L x M, added moles, and divided by 1L which is the same as adding molarities).
pH = 4.27 + log(0.339/0.670) =3.99
Remember to redo this with your own pKa2 and check my work too.
The b part is done the same way but will be slightly more basic. The c and d parts are best worked with an ICE chart and use mols. If you need further help please explain exactly what you don't understand and show work so I can see where you're headed.
for a i got pH=3.97 and for b i got 3.99
for c and d i am stuck.
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