1. (2 pts) Another buffer found in blood is based on the...

1. (2 pts) Another buffer found in blood is based on the equilibrium between dihydrogen phosphate and monohydrogen phosphate. The reaction is shown below: H2PO4^-(aq) + H2O(l) ? H3O^+(aq) + HPO4^2-(aq) If the pH of a blood sample was 7.23, what would you calculate as the ratio of [H2PO4^-] to [HPO4^2-]? (Ka1 = 7.5 × 10^-3, Ka2 = 6.2 × 10^-8, Ka3 = 3.6 × 10^-13) 2. (4 pts) If you needed to perform a reaction in a controlled pH environment that was fairly basic, you might choose to use the ammonium and ammonia buffer system. How many grams of solid ammonium chloride would you have to add to 2.250 liters of 0.180M NH3 to obtain a buffered solution of pH = 9.20? (Kb of NH3 = 1.8 × 10^-5)

Transcript

0:00 Hello.

00:02 So we're looking at the equilibrium where we have h2po4 and hpo4, 2 -minus.

00:09 And i've written out the equations for k -a -1, k -a -2, and k -a -3.

00:13 So you know which one that you needed to grab.

00:15 Now, the equation here that has those two things, h -2 -p -o -4 and h -p -o -4, both, is k -a -2.

00:26 So that means the value will actually be using as going to be the second k -a value.

00:31 Now on top of that, they're interested in the ratio between two of the concentrations of a buffer.

00:39 Naturally, a buffer means we can use the henderson -hosselbach equation, which is not p -k -a, which is ph equals the p -k -a plus the log of the base concentration, which in our case would be the h .p .o .4, because it has one less hydrogen, divided by the acid concentration, which will be the one with an additional hydrogen.

01:05 All right.

01:06 So we know the ph they want us to find.

01:09 That's 7 .23.

01:12 Now the pca, we can get readily enough.

01:15 Pca just equals the negative log of the ka value.

01:20 So that's going to be negative log of 6 .2 times 10 to the negative 8th.

01:28 Throw that into a calculator and you should get 7 .208 for that.

01:34 So we can plug that in.

01:38 And then plus log.

01:39 Now we're actually going to solve for this entire ratio.

01:44 The entire parentheses thing here.

01:52 That whole thing there in red.

01:53 We're solving for that.

01:57 Okay.

01:57 So the first thing we would do is to subtract the 7 .208 from both sides.

02:02 We're trying to just get that fraction by itself.

02:05 Just go down a little bit.

02:07 All right, subtracting that off should give you 0 .0224, which equals log of, and then our whole parenthesis thing is still here.

02:26 Right.

02:26 Now, the way you get rid of a log is you raise it to the, or you do 10th to its power.

02:34 So we're going to do 10.

02:36 And then the whole log, so -and -so is going to be the exponent off the 10th.

02:40 And then we'll do the same thing to the left -hand side.

02:42 So it's 10 to the 0 .024 power.

02:46 This effectively cancels out the log, and all you're left with is the parentheses.

02:52 You will have to plug in the left hand side, so that's 10 to the point 024 power, which should give you 1 .053 in your calculator equals our entire ratio here.

03:10 Now, based on the actual wording of the problem, i think, let's see, what would you calculate as the ratio of h2p -o -4 to hpo4? that wording makes me think they want the h -2 -p -o -4 on top, right, and the h -p -o -4 on bottom.

03:27 To flip that fraction upside down, you're just going to do one over 1 .053.

03:35 That will flip this upside down and give it, i think, the ratio that they're actually asking for.

03:44 One divided by 1 .053 gives you a fraction of 0 .9 .9.

03:49 Which should be your final answer.

03:53 Because it's a ratio, there's no units with it or anything.

03:56 I'm pretty sure that's the answer they're looking for.

03:59 Now you had a second problem on here as well.

04:04 Let's scroll down a bit.

04:05 Start problem number two.

04:09 If you need to perform a reaction in a control, do you need a buffer? how many grams of solid ammonium chloride? ammonium chloride is nh4...

Question

Please give Ace some feedback