15 consider the titration of a 360 ml sample of 0175 m hbr...

15. Consider the titration of a 36.0 mL sample of 0.175 M HBr with 0.205 M KOH. Determine each of the following: a. the pH at 11.1 mL of added base. Express your answer using three decimal places. b. the pH at the equivalence point. Express your answer as a whole number. 11. For 280.0 mL of pure water, calculate the initial pH and the final pH after adding 0.028 mol of NaOH. Express your answers to two decimal places. Enter your answers numerically separated by a comma. For 280.0 mL of a buffer solution that is 0.225 M in HCHO2 and 0.300 M in KCHO2, calculate the initial pH and the final pH after adding 0.028 mol of NaOH. (Ka(HCHO2) = 1.8×10^(-4)). Express your answers to two decimal places. Enter your answers numerically separated by a comma. For 280.0 mL of a buffer solution that is 0.295 M in CH3CH2NH2 and 0.225 M in CH3CH2NH3Cl, calculate the initial pH and the final pH after adding 0.028 mol of NaOH. (Kb(CH3CH2NH2) = 5.6×10^(-4)). Express your answers to two decimal places. Enter your answers numerically separated by a comma. 1. Calculate the molar ratio of NaF to HF required to create a buffer with pH = 4.15. (Ka(HF) = 6.8×10^(-4)). Express your answer to two significant figures.

Transcript

00:01 All right, so we have to consider the 36 .0 ml sample of a strong acid and 0 .205 molar of a strong base.

00:12 Now, to determine each of the following, the first one is when the volume of the base that has been added was 11 .1 ml, so that's 0 .0111, right, basically, liters.

00:30 The number of moles of potassium hydroxide at that point would be 0 .0111 multiplied by its concentration, which is 0 .205.

00:39 That's what we have as its concentration.

00:43 And of course, this would mean that the number of moles was 2 .28 times 10 to the power of minus 3.

00:49 On the other hand, the number of moles of the acid that we would give it would have been 0 .175, which is its concentration, 0 .175, multiplied by its volume, which is 0 .036.

01:03 And what we have is 6 .3 times 10 to the power of minus 3.

01:07 As you can see, the acid is greater.

01:10 The number of moles of the acid is greater than the number of moles of the base, which means that the hydrogen ions are in excess.

01:18 By how much? well, by the difference between the two of them, right? and of course, that would give you 4 .02 times 10 to the power of minus 3.

01:33 And the concentration now of the excess acid is this, 4 .02 times 10 to the power of minus 3, divided by the total volume of the solution, which is this volume plus this volume, right? and that's basically 47 .1.

01:53 So that's 0 .0471.

01:55 And this would give you 0 .0854.

01:59 So the ph at this point is negative log of that.

02:04 And when you punch your calculator, you have 1 .069.

02:08 We're asked to leave our answer in three decimal places, right? okay.

02:12 So the b part, right? when you've added, the b part says, ph at equivalence point.

02:19 The ph at equivalence point is seven, because it's a strong acid and a strong base titration, right? so it's seven.

02:29 It's just that straightforward for that.

02:33 All right.

02:33 Let's look at the next question.

02:35 So the other question that says, for 280 right, for 280 ml of pure water, calculate the initial ph and the final ph after adding 0 .028 moles of sodium hydroxide.

02:57 The initial ph, the ph initial, right? is seven, because it's pure water.

03:07 But once you've added 0 .028 moles of sodium hydroxide, the concentration of the which is what would be 0 .028 divided by the volume, that's 0 .28, right? and this is 0 .1.

03:28 So the ph, right? the ph will be 14 minus minus log of the oh, hydroxide ion concentration.

03:39 And this would be 13 .0.

03:43 When i ask to express, i'll say in two decimal places, right? so 13 .00, right? that's going to be the ph of the solution for that question.

03:57 All right.

03:57 Let's look at the next one.

03:58 Now you have buffer solutions.

04:01 You have buffer solutions.

04:03 So you have 280 ml of the buffer, right? so let's see the concentration of the, the concentration of the formic acid, right? formic acid, hcooh was 0 .225.

04:23 The concentration of the formate ion, that's the sodium formate, was 0 .300 molar.

04:31 This is also molar, right? the ph basically is the pka, the initial ph, plus log the concentration of the base divided by the concentration of the acid.

04:46 And the pka, right? is minus log of the ka.

04:50 Just put that.

04:51 The pka is basically minus log of the ka.

04:55 So, and what you have is the ka was 1 .8 times 10 to the power minus 4.

05:03 So minus log 1 .8 times 10 to the power minus 4 plus the log of 0 .3 divided by 0 .225.

05:13 When you punch your calculator, this will give you 3 .86.

05:18 That is the ph of the solution initially, right? this is initially, initially, right? that's the ph of the solution.

05:30 And the final ph, when you add 0 .028 mole of hydroxide.

05:36 Now on adding, so the concentration of the hydroxide ion, you've added basically 0 .028 divided by the volume of the buffer, which was 280.

05:47 So you've added just 0 .1 molar...