Question

Calculate the ratio of conjugate base to acid at a pH of 8.36.

Answer 1

Recall the fundamental properties of the logarithm
+log(ab)

is the same as

−log(ba)

so Henderson Hasselbach can have plus or minus signs. The correct equation is

pH=pKa+log(conjugate baseacid)

or

pH=pKa−log(acidconjugate base)

Now sort out whether the textbook solution is right or not.
Also;
Important is the knowledge how Ka is defined:

Ka=[H+][A−][HA]

If we perform logarithmization, we get:

logKa=log[H+]+log[A−][HA]

If we apply the operator pX=−logX:

pKa=pH−log[A−][HA]

respectively

pH=pKa+log[A−][HA]

In our case, HA=H2CO3 and A−=HCO3−.

Answer 2

This problem can't be solved until you know the pKa. When you do then
pH = pKa + log [(base)/(acid)] and then you solve for (base)/(acid) to give you the ratio.

Answer 3

@ Helper. I'm confused about what you have written for two reasons.
1. You talk about H2CO3 and HCO3^- but I don't see these compounds/ions anywhere in the problem. Perhaps this is a carry over from a previous problem posted by Steve; however, I looked for that and couldn't find it.
2. The second reason, and perhaps more importantly, is that the division sign (/) has been omitted from each of the equations. They would be correct if the / sign had been included. Technically, they are wrong without the / sign. This is not a big deal since this answer doesn't answer the question but I'm just pointing it out for your information. You may not be aware that the / sign is missing. Have I missed something something?