Khan academy Question 1: practice acid.base equilibria

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Tanphoteric
Posts: 10
Joined: Sat Nov 03, 2018 5:46 pm

Khan academy Question 1: practice acid.base equilibria

Post by Tanphoteric » Sat Nov 09, 2019 3:17 pm

Suppose an equilibrated, dilute solution containing an acid HA with Ka = 10^-4 is measured to have pH = 6 and [HA] = 10^-8 M. What is [A-]?

I get to use henderson hasselbach eq. but idk where the math to separate the log terms for A-/HA comes in? I know that it is simple math but I dont understand how to do it step by step.

acid HA
Ka = 10^-4
pH = 6
[HA] = 10^-8 M.
What is [A-]?

pH = pKa + log [A-] / [HA] *pKa = -log (Ka)

6 = -log(10^-4) + log [A-] / (10^-8) *where do I go from here?
6 = 4 + log [A-] / (10^-8)
NS_Tutor_Mathias
Posts: 282
Joined: Sat Mar 30, 2019 8:39 pm

Re: Khan academy Question 1: practice acid.base equilibria

Post by NS_Tutor_Mathias » Sat Nov 09, 2019 4:44 pm

That looks like you shouldn't need most of the information presented, although you could very well go through the longer way of solving this by using the equilibrium expression from KA and the given concentration of HA. Henderson-Hasselbach is also usable, but is an extremely roundabout way of approaching this.

pH = -log[H+]
-6 = log[H+]
10^-6 = [H+] = [A-]

Since for a weak monoprotic acid, [H+] = [A-]

What in particular gave you trouble with this question? I'm happy to help you get clear up some misconceptions about acid-base chemistry, just let me know what has been giving you trouble.

Edit:
If you really wanted to use H-H, then just take your current last step and simplify, then but both sides to base 10. So:

2 = log([A-]/10^-8)
10^2 = [A-]/10^-8
10^-6 = [A-]

Remember to track your parenthesis correctly, the solution pathway you typed out misplaces them around the logarithm.
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