assignment Exp. 7 Buffers Name: Date: Lab Section: 1. Use the Henderson-Hasselba
ID: 962692 • Letter: A
Question
assignment Exp. 7 Buffers Name: Date: Lab Section: 1. Use the Henderson-Hasselbalch equation to perform the following calculations. The K of acetic acid is 1.8 x 10. Review your calculations with your instructor before preparing the buffer solutions. Buffer A: Calculate the mass of solid sodium acetate required to mix with 100.0 mL of 0.1 M acetic acid to prepare a pH 4 buffer. Record the mass here AND in your data table. Buffer B: Calculate the mass of solid sodium acetate required to mix with 100.0 mL of 1.0 M acetic acid to prepare a pH 4 buffer. Record the mass here AND in your data table. 2. If the Ka for an acid is 5.2 x 106, what is the pK:? 3. If the pK, is 4.50, what is the K.? 63Explanation / Answer
1) I will call acetic acid HAc and acetate Ac-.
pH= pKa + log[Ac-]/[HAc]
pKa= -logKa= 4.74
4= 4.74 + log[Ac-]/0.1M ----> [Ac-]= 0.0182M
[Ac-]= mol/Volume ----> mol= 0.0182M x 0.1L= 1.82x10-3 mol
mass of sodium acetate=1.82x10-3mol x 82,0343g/mol= 0.149g
Buffer B:
pH= pKa + log[Ac-]/[HAc]
4= 4.74 + log[Ac-]/1M
-0.74= log[Ac-]
[Ac-]= 0.182M
[Ac-]= mol/V ---> mol= 0.182M x 0.1L= 0.0182mol
mass of acetate= 0.0182mol x 82,0343g/mol= 1.49g
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2) pKa= -logKa
pKa= -log 5.2x10-6 --> pKa= 5.28
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3) pKa= -logKa= 4.50 ----> Ka= 3.16x10-5
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4) Buffers function best when the pKa of the conjugate weak acid used is close to the desired pH of the buffer. So, if you need to prepare a buffer with pH=3 the pKa of the conjugate acid selected must be close to 3. Let´s find our pair of acid-base to make the buffer:
HClO2 (chlorous acid) --->pKa= -log 1.2x10-2= 1.92
HF (hydrofluoric acid) -----> pKa= -log 7.2x10-4= 3.14 ------> this is the acid selected.
HClO (hypochlorous acid) -----> pKa= -log 3.5x10-8= 7.46
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5) The best method is the one we did in question 4), look at the values of pKa and select the acid with a pKa value close to the desired pH. This is because the maximum buffer capacity is reached when the concentrations of the acid and its conjugate base (or base and conjugate acid) are equal and large. If the concentrations are equal, that means pH=pKa (log1 is 0, look at the Henderson-hasselbalch equation), that is the reason why we search pKa similar to pH, it is to have equal concentrations of acid-base.
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