Posted by **ryan** on Saturday, February 2, 2008 at 11:31pm.

Here are 2 questions I worked out, but need them checked. Also, can't get answer to another question correct unless question 1 is correct. Would you check the first 2 and help me with the third?

1) What is the pH of a 0.05 M solution of TRIS acid (pka = 8.3)?

My answer: I used the formula pH= (pka- log (HA)} /2 then I plugged in:

pH= 8.3-log(0.050)/2

pH= 4.8

2) What is the pH of a 0.045 M solution of TRIS base?

My answer: I used the formula PH=pka+14+log (base)/2

Then i plugged in:

pH=8.3+14+log(0.045M)/2 =

pH=10

3) How many total mL of 1 M NaOH can you add to the solution in problem 1 and still have a good buffer (that is, within 1 pH unit of the pka)?

- chemistry lab -
**drbob222**, Sunday, February 3, 2008 at 12:32am
ryan--Did you post earlier under the name of Matt? I didn't answer that question because a TRIS buffer is made differently. But IF you have a solution of the acid form of TRIS, only (no base present), then 4.80 is the correct answer, I believe.

2. With that same understanding of the problem (no acid present in problem 2), then I found 10.48 for the pH instead of 10. Treat it as you would an ammonia solution (NH3).

RNH2 + H2O ==>RNH3^+ + OH^-

Kb = (RNH3^+)(OH^-)/(RNH2) = 2E-6

(OH^-) = sqrt(RNH2)*2E-6 = sqrt(0.045*2E-6)= 3E-4

pOH = 3.52 and pH = 10.48

Check my thinking. Check my arithmetic.

3. I believe the way to approach this problem is as follows:

RNH3^+ + OH^- ==> RNH2 + H2O

pH = pKa + log (base/acid)

So lets start with 100 mL of the acid. That means we have 0.1 L x 0.05 M = 0.005 mols acid initially. If we add y mols NaOH, we will produce y mols of RNH2 and we will have remaining 0.005-y mols RNH3^+.

9.3 = 8.3 + log(y/0.005-y)

I have y = 0.00455 mols NaOH which is 4.55 mL (which I would round to 4.5 mL) for the volume. That SOUNDS reasonable but check my work. Check my thinking.

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