Use Figure 19.6 to find an indicator for these titrations: (a) 0.10M CH3 NH2 ( Appendix C ) with

Use Figure 19.6 to find an indicator for these titrations:...

Key Concepts

Acid-Base Titration

An acid-base titration is an analytical procedure used to determine the concentration of an unknown acid or base by reacting it with a titrant of known concentration until complete neutralization is achieved. This process involves monitoring the pH change of the solution as the titrant is added, often producing a titration curve that reveals the equivalence point.

Equivalence Point

The equivalence point in a titration is reached when the amount of titrant added is chemically equivalent to the amount of analyte present in the solution. Its pH depends on the strengths of the acids and bases involved; for instance, titrations involving a weak base and a strong acid yield an acidic equivalence point, while those involving a strong acid and a strong base result in a neutral pH at equivalence.

Acid-Base Indicators

Acid-base indicators are substances that change color within specific pH ranges, making them useful for signaling the endpoint of an acid-base titration. The ideal indicator is selected based on its transition range, which should closely match the pH at the equivalence point of the titration.

Indicator Selection in Titrations

The process of choosing an appropriate indicator for a titration depends on predicting the equivalence point's pH. For example, titrations involving a weak base and strong acid generally have an equivalence point in the acidic region, necessitating an indicator that changes color at lower pH values, while titrations between a strong acid and a strong base typically have a neutral equivalence point, requiring an indicator that operates effectively around pH 7.

Transcript

00:01 We're looking for an indicator for the titration of ch3 and h2 with hcl.

00:06 This reaction, since hcl completely associates in water, can also be written as h -plus, or ch3nh2 plus h -plus, goes to ch3nh3 plus, and we don't need the chlorine ions, since they will, they're not, their spectator ions, they don't participate in the reaction.

00:28 In order to find a ph indicator for this titration, we need to know what the ph of the solution will be at the equivalence point, because we want an indicator that will change color at that ph.

00:44 Because hcl is a strong acid, it will react completely with the weak base, ch3 and h2, to form ch3nh3 plus.

00:54 So we can now do this problem as a, or this part of the problem, as a weak acid dissociation problem.

01:05 So ch3nh3 plus plus the water that is in the solution.

01:16 It goes to ch3nh2 plus h3o plus.

01:29 And these are all in the equities phase other than water, which is a liquid.

01:33 And now we can set up our ice table.

01:40 In order to find the initial value for ch3nh3 plus, we need to look at our problem and see how much is formed from the h .h or ch3, nh2, and hcl added at the beginning of the titration, or throughout the titration until we get to the equivalence point.

02:05 So we know we start with 0 .1 molar ch3 and h2, and 0 .1 molar hcl, which also implies 0 .1 molar h plus ions.

02:17 There's an equal amount of ch3 and h2 and h plus, and there is a one -to -one ratio in the reaction between them.

02:27 So all of the ch3 and h2 and h -plus that we have will be completely reacted.

02:34 So whatever number of moles we have for other one of these, which are equal, are going to also equal the number of moles of reactants we get out after the reaction.

02:50 But since we are also adding these two, the volume will double.

02:55 And since we have an equal number of moles, but a doubled volume, our molarity will have.

03:03 So we'll end up with 0 .05 molar ch3 and h3 plus.

03:10 So now we know our initial value here, 0 .05 molar.

03:17 And then we start with 0 molar ch3 and h2 and 0 molar h3o plus.

03:24 And liquids are included in these calculations, so we can just ignore water here.

03:32 For the change, we react some unknown amount that we can call x to form x.

03:38 X of each product and we'll solve for x later.

03:45 Our equivalence will be our initial amount plus our change.

03:49 So for our ch3 nh nh3 plus it is 0 .05 minus x.

03:55 For the product it's just x for each of them.

04:01 I just realized i made a mistake here.

04:03 This should be just kb, not pkb.

04:06 It's an easy mistake to make, but it really messes up with your calculations...

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