Acetic Acid=-log(1.8x10^-5)= 4.75 Phosphoric acid (H3PO4) -logKa= pKa -log(7.2x1
ID: 510468 • Letter: A
Question
Acetic Acid=-log(1.8x10^-5)= 4.75
Phosphoric acid (H3PO4) -logKa= pKa -log(7.2x10^-3) = 2.14 -logKa= pKa -log(6.3x10^-8)= 7.2 -logKa=pKa -log(4.2 10^-13)= 12.4
21. If you want to prepare a 0.1M buffer for pH 4.5, which of the above acids would be the better choice for the buffer? Explain. pH = pKa + log ([base conjugate]/[acid]) Or Ka= 10 ^-pH Ka= 10 ^-4.5= 3.16x10^-5, so use acetic acid because it is closest.
22. What salt (sodium salt of the conjugate base) and acid would be used to prepare the buffer? Acetic acid, and sodium acetate.
23. Write the acid dissociation equation including the conjugate base: (is this right? If not, please help!) CH3COOH(aq) --> CH3COO(aq) + H+(aq)
24. Assuming pH 4.5 is on the acidic side of the equivalence point, calculate how to prepare 1L of a 0.1M buffer with pH 4.5 for each of the above two systems (calculate amounts of acid and conjugate base for each systems in moles).
25. Do the molar ratios of acid to conjugatebase in each buffer system confirm your choice in question 3 above? or refute it? Provide an explanation of why your choice would make a good buffer?
Explanation / Answer
21. To prepare a buffer we use weak acid or weak base whose pKa is closest to the pH we requrie for the final buffer solution.
Thus to prepare a buffer of pH 4.5, we would use acetic acid/sodium acetate as the wek acid/conjugate base system
22. We would use acetic acid and sodium acetate as the buffer components.
23. to prepare buffer of pH 4.5
Using Hendersen-Hasselbalck equation,
pH = pKa + log(base/acid)
= pKa + log(sodium acetate/acetic acid)
So,
4.5 = 4.75 + log(sodium acetate/acetic acid)
(sodium acetate) = 0.56(acetic acid)
we have
(acetic acid) + (sodium acetate) = 0.1 M x 1 L = 0.1 mol
(acetic acid) + 0.56(acetic acid) = 0.1 mol
(acetic acid) moles = 0.1/1.56 = 0.064 mol
mass of acetic acid required = 0.064 x 60.05 = 3.84 g
(sodium acetate) moles = 0.1 - 0.064 = 0.036 mol
mass of soium acetate required = 0.036 x 82.03 = 2.95 g
25. molar ratio of (base/acid) = (0.036/0.064) = 0.5625
which when fed to the Hendersen-Haselbalck equation
gives pH = 4.5
thus it is good combination to prepare the needed buffer