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1) If you added 3 ml of 1 M NAOH to your 100ml of buffer,would it still be usable buffer according to our conventions?Explain why or why not(use math).you may do this problem on one of your...

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1) If you added 3 ml of 1 M NAOH to your 100ml of buffer,would it still be usable buffer according to our conventions?Explain why or why not(use math).you may do this problem on one of your buffers.2)Calculate the theoretical ph of one of your buffer at 0c.Assume that the room temperature is 22c,if none of your buffers was listed on the table on thechangingpka with temperature do this problem for TRIS at ph 8.03)what would be the most efficient way to make up a HEPES buffer at ph 8.5?what starting compound and reagents would you choose to use?4)When Dr. Farrell was a graduate student, he once made up a pH 8.0 sodium acetate buffer. Why would the casual observer to this buffering faux pas come tot eh conclusion that he had the intellectual agility of a small soap dish?5)if you made up a solution of 100 ml 0.1 M TRIS in the acid form, what would be the ph?6)if you added 3ml of 1MNAOH to the solution 5 what would be the ph?
weak acid ph=pka-log(HA)/2weak base ph=pka+14+log(A-)/2ph=pka+log(K-/HA)
Answered Same Day Dec 20, 2021

Solution

David answered on Dec 20 2021
128 Votes
1. If you added 3 ml of 1 M NAOH to your 100ml of buffer,would it still be usable buffer
according to our conventions? Explain why or why not(use math).you may do this problem on
one of your buffers.
Concentration of the buffer = 0.05 M
Volume o the buffer = 100 ml
PH of the buffer = 7.2
pKa o phosphate buffer [from the literature] = 7.5
We are asked to find out the resultant PH when 15 ml of 0.01 M NaOH and the concentrations of
H2PO4
-
and HPO4
2-
in this final solution.
The dissociation of phosphoric acid proceeds as follows:
H3PO4 <----- [H+ +H2PO4
-
] ------ [H
+
+ HPO4
2-
] ------------ [H
+
+ PO4
3-
]
pH = pKa + log [A-]
[HA]
Where
[A
-
] = [HPO4
2 -
]
[HA] = [H2PO4
-
]
log 10 [HPO4
2 -
] = pH - pKa
[H2PO
4-
]
[HPO4 2
-
] = 10 (pH- pKa ) = 10(7.5-7.2) = 2.0
[H2PO4
-
]
[HPO4
2-
] +[H2PO4
-
] = 0.1 M so,
[H2PO4
-
] = 0.0333 M and [HPO4
2-
] = 0.0667 M
When 3 ml of 1 M NaOH is added to 100 mL of 0.05 M phosphate buffer,
Molarity of NaOH * Volume of the solution = Number of moles of NaOH
[1 mole/litre] * [0.003 M] = 0.003 moles of NaOH
0.003 moles of NaOH on hydrolysis gives equivalent 0.003 moles of OH
-
ions. These combine
with 0.003 moles of protons [obtained from dissociation of phosphoric acid] to form water.
HPO4 3

HPO4 2



So, the resulting concentration [number of moles] of [HPO4
2-
] would be
= original number - 0.003
= 0.0333 - 0.003
= 0.0303
pH = pKa + log10 [A-]
[HA]
= 7.5 + log [0.0303]
[0.0677]
= 7.5 + log (0.4475)
= 7.5 – 0.3491
= 7.15
So, the resulting PH of the phosphate buffer when 15 ml of 0.01M NaOH is added = 7.15
While PH of the...
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