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September 18, 2014

September 18, 2014

Posted by **Elizabeth** on Wednesday, November 10, 2010 at 4:51pm.

- Please help -
**Elizabeth**, Wednesday, November 10, 2010 at 6:38pmNeed answer tomorrow for pre-lab Thanks

- Chemistry -
**DrBob222**, Wednesday, November 10, 2010 at 8:54pmUse the Henderson-Hasselbalch equation.

pH = pKa + log (base/acid)

pKb = 4.70; therefore, pKa = 14.0-4.70 = 9.3

Solve two equations simultaneously.

Plug in 9.0 for pH and solve for (B/A) and I get 0.501 which you should confirm. If B/A = 0.501 then

(Base) = 0.501*(acid)

The second equation is

(base) + (acid) = 0.05

Solve for base and acid. I get something like

(acid) = 0.0333 M

(base) = 0.0167 M

We have two solutions, both 0.1 M with which to make this. Use a dilution process (formula) to determine how much of each should be taken.

0.1M x (x Liters/0.25 L) = 0.0333 M.

Solve for x = 0.08325 L or 83.25 mL acid.

0.1 M x (x liters/0.25) = 0.0167 M

Solve for x = 0.04175 L or 41.75 mL base.

Therefore, the instructions would be to measure 83.25 mL of NH4Cl and 41.75 mL NH3, place in a 250 mL volumetric flask and make to the mark with water. But you shouldn't stop there. You should go through and prove four things.

1. The final solution is 250 mL.

2. The final solution is 0.05 M.

3. The final (acid) is 0.0333 (or whatever value you find above) and the final (base) = 0.0167 (or whatever value you find above.)

4. The final pH is 9.0.

- Chemistry -
**Elizabeth**, Wednesday, November 10, 2010 at 10:14pmThanks!

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