Estimating Ka

[C5b6789]

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In the TPR Science Workbook explanations for GChem, they estimate the value of Ka but I'm a confused on how they are doing so.

Passage 64 #1
What is the approximate pKa for HF? { Given in passage: Ka for HF is 6.8 x 10^(-4) }

Answer: Since the Ka for HF is 6.8 x 10^(-4), which is between 10^(-4) and 10^(-3) etc...

I don't understand how it's between 10^(-4) and 10^(-3) when...

0.001 = 10^(-3)
0.0001 = 10^(-4)
0.00068 = 6.8 x 10^(-4)


Another example (of many): Passage 68 #5

"Since 1.5 x 10^-2 is between 10^-1 and 10^-2..."

0.1 = 10^(-1)
0.01 = 10^(-2)
0.015 = 1.5 x 10^(-2)


I'm not very good at math so I think I'm confused about something straightforward like decimal places and exponents...apologies if this is a dumb question.

 

[gettheleadout]

No dumb questions! It does appear that you're getting thrown off by the scientific notation. Let's take a look.

Passage 64 #1: The Ka for HF is given in the passage as 6.8 x 10^–4 right? So if we look at that, we see that if we wanted to work with nicer, round numbers in our heads, we could round the 6.8 up to 10, in which case we would have 10 x 10^–4, which you should recognize would simply be 1 x 10^–3 which itself is just 10^–3. Our other option is to round the 6.8 down to 1, giving us 1 x 10^–4 which itself is just 10^–4. The three possible values we have to look at are then:

6.8 x 10^–4 (the original given)
1 x 10^–3 (after rounding up)
1 x 10^–4 (after rounding down)

Since the latter two values resulted from rounding the original in opposite directions, the original must be a value between the two. So, 6.8 x 10^–4 is the middle value. Relative to the original given, we got 1 x 10^–3 by rounding up, so its value is greater. Then, we got 1 x 10^–4 by rounding down, so its value is less. This gives the inequality:

(1 x 10^–4) < (6.8 x 10^–4) < 1 x 10^–3

This is a relationship that should become apparent very easily as you become comfortable with scientific notation. See if you can apply the same logic to your second example.

From this point, if we are trying to estimate pKa from the given Ka, there are two options. From the answer explanation excerpt you provided, I can surmise that TPR is doing it using what I might call the range method, saying "look, the Ka we have is between 1 x 10^-3 and 1 x 10^-4. The pKa's for those values would be 3 and 4 respectively, so the pKa for HF is between 3 and 4." This is a valid method, and assumes knowledge of the following relationship: – log (1 x 10^–q) = q

That method may be sufficient to get an answer. The other option is a method described in TBR Chemistry, using a mathematical identity I like to call the z-y log rearrangement. The relationship is thus: – log (z x 10^–y) = y - log (z). Applying this relationship to the given Ka reveals that pKa = – log (6.8 x 10^–4) = 4 - log (6.8)

Computing pKa from that point requires estimation of the value of "log (6.8)." For this method the values of the logs of common small integers is recommended. If one knows that log (7) = ~ 0.85, then we can estimate that 4 - log (6.8) = 4 - (~ 0.8) = ~ 3.2. As you can see, this method, while more intensive, can yield a much greater degree of accuracy than the range method.

 

[C5b6789]

Thank you so much for your explanation. I was actually using the "z-y log rearrangement" since that's how the Math Review section of TPR's Physics Review Book explained it, but for some reason the SWB answers use the range method. I think I'll stick to the z-y log rearrangement since it makes more sense to me! Thanks again 🙂