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Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: 05holtel on February 08, 2009, 12:35:58 PM

Title: Calculation when PH=PKA
Post by: 05holtel on February 08, 2009, 12:35:58 PM

if 50 ml of 0.01M NaOH is added to 100 ml of a 0.05M phosphate buffer at pH 7.2. What is the resultant pH and the resultant concentrations

pH =pKa + log[HPO42-]/H2PO4-]
   = 7.2+ log (0.0025+ 0.0005)/ (0.0025-0.0005)
   pH= 7.38

Apparently,
[H2PO4-] =(0.0025-0.005) mole/100ml +50 ml  x1000ml/1L= 0.0133M
[HPO42-] = (0.0025+0.005)  mole/ 100ml+ 50 ml x 1000ml/1L= 0.0200M

I am not sure how they got those concentration values. When I plug the numbers into my calculator, I get different values.

PLease Help

Title: Re: Calculation when PH=PKA
Post by: enahs on February 08, 2009, 01:16:07 PM

I get those numbers when I do the math.

Remember, you are starting at a pH of 7.2, which is the pKa point of the buffer system. That means even though you have 0.1L * 0.05 M = 0.005 moles of buffer solution, you are at the pKa point, and thus half is in one form (acidic) and half is in the other form (basic).