Use Figure 19.6 to find an indicator for these titrations: (a) 0.10M HCl with 0.10M NaOH (b) 0.1

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

Key Concepts

Acid-Base Titration

Acid-base titrations involve gradually adding a base to an acid (or vice versa) to reach a point of complete neutralization. The process is represented by a titration curve that displays the pH change as the titrant is added, helping to determine the equivalence point where the amounts of acid and base have stoichiometrically reacted.

Equivalence Point

The equivalence point is the stage in a titration at which the quantity of titrant equals the amount required to completely neutralize the analyte. Its pH depends on the strength of the acid and base; for instance, titrations of a strong acid with a strong base typically have an equivalence point near pH 7, while weak acid–strong base titrations result in an equivalence point in the basic range.

Indicator Selection

Indicator selection is based on choosing a substance that changes color in the pH range close to the equivalence point of the titration. The correct indicator will undergo a distinct color transition during the rapid pH change near the equivalence point, ensuring accurate detection of the endpoint.

Titration Curve Analysis

Titration curve analysis involves studying the graph of pH versus titrant volume to understand the neutralization process. It reveals the pH jump near the equivalence point and assists in selecting a suitable indicator by matching its transition range to the steep portion of the curve.

Transcript

00:06 In this problem, we are trying to find our indicator for each of the following titrations.

00:18 We know that indicators will have a ph value that is approximated by pka plus or minus 1.

00:27 For part a, we are given 0 .1 molarity of hclp, which is a strong.

00:36 Acid and 0 .1 molars of n .a .o .h which is a strong base.

00:44 A titration between a strong acid and a strong base will result in an equivalence point with the ph equals to 7.

01:04 And by looking at the indicator chart when our ph is equal to 7, this would give us bromothymol blue.

01:20 In part b, we are given 0 .1 molars of h .c .o .h, which is a strong, sorry, a weak acid, and 0 .1 molars of n .a .oh .h, which is a strong base.

01:38 With a titration between a weak acid and a strong base, we will have an equivalence point of a ph that is greater than 7.

01:52 So we will need to start by constructing an ice table.

02:07 Given here is a reaction equation that we will write between the weak acid and strong base.

02:20 And to keep in mind, at an equivalence point, our number of moles of acid and our number of moles of base is equal to each other.

02:30 So as we solve for our ice table with the following reaction, our initial number of moles would be the same.

02:44 We will have 0 .1 moles of h -c -o -o -h and 0 .1 moles of n -a -o -h.

02:56 On the right side, we will start with an initial moles of 0 because the reaction has not proceeded yet.

03:09 Next would be our change.

03:13 Since the limiting reagent is both, they are both equivalent, we will minus 0 .1 moles on both sides, on the left side, and add 0 .1 moles on the right side, side.

03:31 This will give us a result of 0 .1 moles of c .o .h.

03:42 Next, we will need to find the malarity of c .o .h.

03:51 Because both the acid and the base takes up 1 liter, where h.

04:01 1 liter and n -a -o -h is another liter, the addition of the two will give us 2 liters in volume.

04:09 So when we solve for the molarity of c -o -o -h, it'll be our number of moles, which is 0 .10 moles over 2 liters, which gives us our malarity of 0 .050 malarity...

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