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Chemistry Forums for Students => Analytical Chemistry Forum => Topic started by: tashkent on October 14, 2004, 10:25:20 AM

Title: millimole question
Post by: tashkent on October 14, 2004, 10:25:20 AM

Greetings!

I need help on this problem found in #114 of the GRE Chemistry Practice Book:

H2PO4- = HPO4-2 + H+  Ka = 5 x 10-8

114) Given the information shown above, how many millimoles of K2HPO4 must be added to 100 mL of a 0.100 M KH2PO4 solution to obtain a solution with a pH of 7.0?

a) 1.0 mmol
b) 5.0 mmol
c) 10 mmol
d) 20 mmol
e) 25 mmol

I want to use the Henderson Hasselbalch equation pH= pka + log /[HA], but the GRE exam does not allow the use of calculator.  Besides, its so hard to know the pKa if you will not use one (pka is 7.3 if you will use a calculator).  Are there any strategies in solving this problem manually?  The answer for the problem above is B.

Hope to hear suggestions.  Thanks!

Sincerely,
Tashkent

Title: Re:millimole question
Post by: Demotivator on October 14, 2004, 02:26:28 PM

Don't need henderson equation. Just equilibrium equation

K = [H+][HPO4]/[H2PO4]
now plug in the numbers. PH 7 means [H+] = 10^-7.
Solve for moles/liter HPO4, then convert to milimoles given 100 ml.

Title: Re:millimole question
Post by: AWK on October 15, 2004, 03:20:38 AM

[H3O(+)] = K x [H2PO4(-)]/[HPO4(2-)]
10^-7 = 5 x 10^-8 x [H2PO4(-)]/[HPO4(2-)]
2 = [H2PO4(-)]/[HPO4(2-)]
In buffer solution, instead of concentration
you can use moles of reactants (also milimoles) -WHY !?

100 mL of a 0.100 M KH2PO4 contains 100 x 0.1 = 10 milimoles of KH2PO4
2 = [10/[milimoles_of_HPO4(2-)]
hence milimoles_of_HPO4(2-) = 20

My fault n should be 5. Thanks to haiph12 for pointing out.

Title: Re:millimole question
Post by: haiph12 on October 17, 2004, 11:29:59 AM

2 = [ 10/[milimoles of K2HPO4]]
hence milimoles of K2HPO4 = 5

the answer for the problem is B