consider the titration of 500 of 0100 m hcl with 0200 m naoh calculate the at initial stage choi

1. Initial stage: We have 50.0 mL of 0.100 M HCl, and no NaOH has been added yet. To find the in

Consider the titration of 500 of 0100 m hcl with 0200 m naoh...

Consider the titration of 50.0 mL of 0.100 M HCl with 0.200 M NaOH. Calculate the pH at the initial stage. Choices: a) pH = 12.00 b) pH = 4.00 c) pH = 1.00 d) pH = 8.00 Calculate the pH after the addition of 10.0 mL NaOH. Choices: a) pH = 10.40 b) pH = 1.30 c) pH = 6.59 d) pH = 3.60 Calculate the pH after the addition of 20.0 mL NaOH. Choices: a) pH = 1.85 b) pH = 6.53 c) pH = 2.13 d) pH = 1.96 Calculate the pH after the addition of 25.0 mL NaOH. Choices: a) pH = 7.00 b) pH = 2.00 c) pH = 11.00 d) pH = 9.00 Calculate the pH after the addition of 30.0 mL NaOH. Choices: a) pH = 1.90 b) pH = 12.10 c) pH = 13.10 d) pH = 0.90

Transcript

00:01 Hi, so in this case we're going to have the titration of 50 milliliters of a 0 .1 molar solution of hcl.

00:11 We are going to have the titration using a 0 .2 molar solution of sodium hydroxide and we are asked to get the ph at different points.

00:23 So first of all remember that this is going to be a one -to -one relation in our neutralization reaction.

00:36 Of course, sodium chloride is going to appear as ion into solution and actually the reactive species are the hydronium ion and the oh that are going to react.

00:47 So the different cases for which we need to calculate ph are the following.

00:55 The initial, after we have added 10 milliliters of naoh, when we added 20 milliliters, 25 milliliters and 30 milliliters of solution.

01:16 So what i'm going to do is take an advantage that we have a one -to -one relationship.

01:23 Let's do a table with the amount of mole that we have added at the three different points.

01:30 So let's say for case one we have the volume added, we are going to have the amount of mole of oh - that we added, we're going to have the initial amount of mole of hcl which corresponds to the amount of mole of h3o+.

01:57 These values are going to stay the same for all of the cases.

02:07 And then we're going to have the surviving species, the concentration and the value for ph.

02:21 So let's begin by 0, 10, 20, 25 and 30 milliliters.

02:28 So how can i get the amount of mole of oh - that i have added? simply i need to multiply the volume in liters.

02:40 So we can convert all of these to liters simply by multiplying by times 10 to the minus 3.

02:45 We multiply this by the molarity of my solution and the molarity of my solution is going to be 0 .2 mole.

03:00 So the amount of mole that we're going to get is 2 times 10 to the minus 3 mole, 4 times 10 to the minus 3 mole, 5 times 10 to the minus 3 mole and 6 times 10 to the minus 3 mole.

03:20 We can do the same thing with hcl but that initial amount is going to be the same always.

03:27 So n is 50 times 10 to the minus 3 liters because we have 50 milliliters of solution times 0 .1 mole molar concentration.

03:39 So the amount is going to be 5 times 10 to the minus 3 mole.

03:49 Okay so i'm going to modify this table a little bit just to make room for some notes.

04:01 We have the surviving species, amount of mole of the surviving species which is going to be either hcl or oh - and then the concentration of that surviving species.

04:14 With that we can calculate the p.

04:16 Okay so let's begin with the very first case.

04:22 We're going to have volume added 0 liters, the surviving species it is going to be equal to the hydronium ion that is in our solution.

04:31 The amount of mole that we have is 5 times 10 to the minus 3 of h2o plus and the concentration is the same as the concentration of the solution.

04:42 So the concentration is going to be 0 .1 molar and the ph is directly the minus logarithm of that concentration.

04:51 So the ph when we have added nothing it's going to be equal to 1 for my initial solution.

04:59 Next up what i want you to see is that for these two cases the limiting reagent is going to be oh-.

05:11 That means that the species that is going to survive is h2o plus...

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