I just don't get how to find the pH of a solution

[helpmeonchemistry]

Hi,

I'm having extreme difficulty understanding this topic. I'm confused on when I'm supposed to use henderson-hasselbach equation and when to use BCA and ICE Charts.

For example, this problem:

What is the pH of a solution that results when 0.010 mol HNO3 is added to 500 mL of a solution that is 0.10 M in aqueous ammonia and 0.20 M in ammonium nitrate. Assume no volume change. ( The Kb for NH3 = 1.8 x 10^-5)


I know how to solve the problem, but I just don't get why I do what I do. Why do I use set up a BCA chart and why do I set up an ICE chart. Also why is it that I cannot use Henderson-Hasselbach on this.

If anybody can link me to a good resource for this topic, that'd be cool

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[BerkReviewTeach]

First, you should always analyze what you have. In this case you have a strong acid (HNO3) and a buffer (mixture of a weak base, NH3, with its conjugate acid, NH4+), so there will be some type of reaction taking place.

Next, you should consider the quantities. The solution has 0.05 moles NH3 (0.5 L x 0.10 M) and 0.10 moles NH4+ (0.5 L x 0.20 M). The addition of 0.01 moles of a strong acid will convert 0.01 moles NH3 into NH4+, so after reaction you have 0.04 moles NH3 and 0.11 moles NH4+. This is still a buffer, so the HH equation for buffers is needed: pH = pKa + log [NH3]/[NH4+]. It's a matter of plugging in at this point (and not using a calculator, as you do not get that luxury on the MCAT.)

pKa = 5 - log 1.8 = 4.74

so, pH = 4.74 + log (0.04).0.11) = 4.74 + log 4/11 ≈ 4.74 + log 1/3 = 4.74 + (-0.48) = 4.74 - 0.48 = 4.26. The actual answer should be about 4.28.

Now had this been an MCAT question, I would have suggested the following. The buffer starts at pH = 4.74 + log (0.10/0.20) = 4.74 + log 1/2 = 4.74 - log 2 = 4.44. Adding acid to a buffer lowers the pH slightly, so the best answer is a little less than 4.44.

So had this question come with answer choices like:
A) 5.46
B) 5.20
C) 4.28
D) 4.02

Then the best answer would have been the one closest to, but less than 4.44 (choice C).

 

[NextStepTutor_1]

Hi @helpmeonchemistry -

Just some high-level points that may also be helpful...it is indeed the case, as @BerkReviewTeach pointed out, that all pH calculations have to be based on what we have. ICE/BCA tables are useful because they are ways to systematically figure out what we actually have once we account for the conditions presented in the question.

Here's an analogy that may help: imagine that you will receive a payment of 1% interest on whatever is in your savings account tomorrow. You currently have $120 in the account, but you will have to withdraw $20 today to cover some other expenses. How much interest will you receive tomorrow? Well, tomorrow you will have $120 - $20 = $100, and $100 * 0.01 = $1, so you will receive $1 in interest. This example is pretty straightforward, but bear with me -- it's actually quite similar to buffer problems like the one you asked about. Ka (or its related quantity, pKa) is like the 1% interest in that analogy, because Ka is basically a measure of the proportion of an acid that will dissociate. The final [H+] concentration is kind of like the final payment you will receive, and the calculation of $120 - $20 = $100 is like how you have to account for the changes that happen when you put an acid/base into the buffer solution.

The basic difference is that the word problem above is easy -- the circumstances are pretty straightforward (you have to pay some money before you get your interest payment) and the calculation of $120 - $20 = $100 is easy. Acid-base chemistry is harder. You've got to deal with multiple compounds with different types of chemical behavior, as well as scientific notation, logarithms, and various equations. ICE/BCA tables basically serve as a systematic way to organize your thoughts as you figure out what you actually have in the solution, which is what you need for pH calculations.

Returning to pH calculations—which again, are based on what you actually have in the solution—don't lose sight of the fact that the definition of pH is fairly simple: pH = -log[H+], so if we know [H+], we know the pH. There are a few common ways you will see of figuring out [H+]. Sometimes you can just use the concentration of the acid and assume that it will completely dissociate, like with HCl. Sometimes you have to fiddle around with Ka, which will always have a term for [H+], because Ka = [H+][A-]/[HA]. You often will do this with weak acid equilibrium problems that involve ICE tables, because Ka is a special case of an equilibrium constant. The Henderson-Hasselbach equation is yet another tool you can use to figure out [H+].

Hope this helps & best of luck!

 

[9bbn]

Hi @helpmeonchemistry -

The final [H+] concentration is kind of like the final payment you will receive, and the calculation of $120 - $20 = $100 is like how you have to account for the changes that happen when you put an acid/base into the buffer solution.

Love this analogy!! @NextStepTutor_1