-Now you have your 1.0 M NaH2PO4, and the 1.0 M Na2HPO4, 1 M HCl, 1 M KOH made b
ID: 1024809 • Letter: #
Question
-Now you have your 1.0 M NaH2PO4, and the 1.0 M Na2HPO4, 1 M HCl, 1 M KOH made by others in the class, in addition to ultra pure H2O. But your pH meter is not working! Describe how you could use only these available reagents to make a buffer that is 225 mM phosphate at pH 6.5. (5 pts)
-You carry out a reaction in 20 mL of the buffer described in the first part . During the course of the reaction 10 mM (millimolar) base is produced. What will be the resultant pH of the solution upon completion of the reaction?
Explanation / Answer
phosphate has a pKa of 7.21
pH = pKa + log base/acid
6.5 = 7.21 + log base/acid
log base/acid = -0.71
base/acid = 10-0.71
base/acid = 0.195
total phosphate is 225 mM
base + acid = 225 x 10-3 M
base = 225 x 10-3 -acid or 0.225 -acid
0.225-acid/acid = 0.195
acid = 0.225/1.195
acid = 0.188 M and base = 0.225-0.188 = 0.037 M
So we will take 188 mL of 1.0 M NaH2PO4 and 37 mL of 1.0 M Na2HPO4 and make the solution to a total of 1 L by adding 775 mL of pure H2O
If you generate 10 mM of base
you base concentration is 0.047 M and your acid concentration reduces to 0.178 M
so
pH = pKa + log base/acid
pH = 7.21 + log 0.047/0.178
pH = 7.21 - 0.578
pH = 6.63
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