What is the most prevalent species at pH = 10? A. At what pH do you have ~99% B. If you have access to both 10 M KOH and 10 M HCl, how much of which would you add to change 1 L of a 100 mM solution of the buffer described by this equilibrium from 3 to 4? C. If you started at pH 8 and added 250 μL of HCl, to the 100 mM solution, what would be the final pH?

Question

Chemistry

Woofie

20 January, 14:37

What is the most prevalent species at pH = 10? A. At what pH do you have ~99% B. If you have access to both 10 M KOH and 10 M HCl, how much of which would you add to change 1 L of a 100 mM solution of the buffer described by this equilibrium from 3 to 4? C. If you started at pH 8 and added 250 μL of HCl, to the 100 mM solution, what would be the final pH?

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Answers (1)

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Marcel Franklin · Answer 1

Explanation:Firstly you need to do the calculations via Handerson Hasselbach equation:

pKa's of phosphoric acid are 2.3, 7.21 and 12.35. If required pH is 6, then, 7.21 will be used. This means, monopotassium dihydrogen phosphate and dipotassium monohydrogen phosphate (diprotic (H2PO4-) and monoprotic (HPO4--) potassium salts) will be used.

pH = pKa + log ([A-]/[HA])

Now, for A - you put K2HPO4 concentration, and for HA you put KH2PO4 concentration:

6 = 7.21 + log (A/HA)

log (A/HA) = -1.2

A/HA = 0.063

So, now you know the fold difference between these salts, and the Molarity of the solution would be given to you, and you can calculate the required amount:

HA+A=0.05

A/HA=0.063

HA=0.047M

A=0.003M

This means, you need to put 0.047 moles of KH2PO4 and 0.003 moles of K2HPO4 salts into 1 liters of solution.

for molecular weights: KH2PO4 = 39+97=136amu; K2HPO4 = 39*2+96=174amu

Thus, 0.047*136=6.392g of KH2PO4 and 0.003*174 = 0.522g of K2HPO4 should be added.

But, you should also note that, if |pH-pKa| >1, then buffer capacity of solution decreases. In this case, it is equal to 1.2