You were asked to prepare a buffer with a pH of 9.00. Table 1: Weak Acid Ka pKa
ID: 546108 • Letter: Y
Question
You were asked to prepare a buffer with a pH of 9.00.
Table 1:
Weak Acid
Ka
pKa
Lactic Acid (HC3H5O3)
1.4 x 10-4
3.85
Acetic Acid (HC2H3O2)
1.8 x 10-5
4.74
Carbonic Acid (H2CO3)
4.4 x 10-7
6.36
Dihydrogen (H2PO4-)
6.2 x 10-8
7.21
Ammonium ion (NH4+)
5.6 x 10-10
9.25
Hydrogen Carbonate (HCO3+)
4.7 x 10-11
10.32
A. Using Table 1, what buffer system will you use?
B. Your lab supervisor asked you to use 0.050 mol of the weak acid component in preparing the above-mentioned buffer. Calculate the amount of (mol) of the conjugate base that you need in order to prepare the buffer.
C. If you were required to use 0.300 mol of the weak base component, how many moles of the weak acid component will you need to prepare the buffer?
Weak Acid
Ka
pKa
Lactic Acid (HC3H5O3)
1.4 x 10-4
3.85
Acetic Acid (HC2H3O2)
1.8 x 10-5
4.74
Carbonic Acid (H2CO3)
4.4 x 10-7
6.36
Dihydrogen (H2PO4-)
6.2 x 10-8
7.21
Ammonium ion (NH4+)
5.6 x 10-10
9.25
Hydrogen Carbonate (HCO3+)
4.7 x 10-11
10.32
Explanation / Answer
Part A
The buffer solutions work best when they have pH = pKa + or - 1
Thus if we need to prepare a buffer of pH = 9.0 , we need to choose an acid with pKa around 9.0, which is ammonium ion here.
Thus I would choose to prepare an ammoium ion, ammonia buffer.
Part B
The pH of buffer is calculated using Hendersen equation as
pH = pKa + log [conjugate base]/[acid]
Given [acid] = 0.05 mol and pka = 9.25 . let us assume x is the number of moles of conjugate base
then
9.00 = 9.25 + log( x/ 0.05 )
Thus solving for x , we get x= 0.0281 mol
Thus the moles of base component required =0.0281 mol
Part C
If [conjugate base ] = 0.300mol , then
9.00 = 9.25 + log (0.3)/x
and x = 0.5335 mol
tha moles of weak acid needed to make buffer = 0.5335 mol
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