Find the pH during the titration of 20.00 mL of 0.1000 M triethylamine, (CH3 CH2)3 N(Kb=5.2 ×10^

Find the pH during the titration of 20.00 mL of 0.1000 M...

Key Concepts

Acid-Base Titration

Acid-base titration is a method used to determine the concentration of an analyte by reacting it with a titrant of known concentration. This process involves gradual addition of the titrant to the analyte solution until the reaction reaches a specific point, often indicated by a change in pH. The titration in this context involves a weak base and a strong acid, requiring careful monitoring of pH changes as the conversion occurs.

Buffer Solutions

Buffer solutions are mixtures that resist changes in pH when small amounts of acid or base are added. In a titration, when the amounts of a weak base and its conjugate acid are present, the solution acts as a buffer. The titration process passes through a buffering region where the pH can be calculated using buffer equations, making it relatively stable against additions of titrant.

Henderson-Hasselbalch Equation

The Henderson-Hasselbalch equation is a practical relationship used to estimate the pH of a buffer solution. It relates the pH to the pKa (the acid dissociation constant of the conjugate acid) and the ratio of the concentrations of the base and conjugate acid. This equation is particularly useful in the buffer region of an acid-base titration to determine the pH after partial neutralization.

Conjugate Acid-Base Equilibrium

In acid-base chemistry, the conjugate acid-base pair plays a central role in defining the chemical behavior of species in solution. When a weak base reacts with a strong acid, it forms its conjugate acid. The equilibrium between the weak base and its conjugate acid involves their dissociation constants (Kb and Ka respectively), which are interrelated through the ion product of water (Kw). Understanding this equilibrium is essential for calculating the pH at various points in the titration.

Equivalence Point

The equivalence point in a titration is reached when the amount of titrant added is stoichiometrically equivalent to the amount of analyte present. In the titration of a weak base with a strong acid, this point is marked by the complete conversion of the weak base into its conjugate acid. The pH at the equivalence point is determined by the hydrolysis of the conjugate acid, and it is typically less than 7 due to the acidic nature of the conjugate acid.

Transcript

00:01 Question number 57 is a weak -base strong acid titration problem, where you are given eight different volumes during the titration process and asked to calculate the ph at each of those eight volumes.

00:20 To do this, we need to recognize that when we start with a weak base in solution, ph is determined using a kb weak -base calculation.

00:33 Then once the titration begins and strong acid has been added, some of the strong acid will convert the weak base into weak acid and we will have a buffer solution.

00:44 So pre -equivalence, all ph calculations will be buffer solution calculations using the henderson -hasselbalch equation.

00:53 At the equivalence point, all of the weak base will have become weak acid, and we will perform a ca or weak acid ph calculation.

01:02 Then post equivalence, we will have excess strong acid, and we will determine the ph simply based upon the excess strong acid concentration or the excess hydronium concentration.

01:18 So let's get started.

01:20 The first volume of strong acid added is zero milliliters.

01:24 If we have added zero milliliters of strong acid, then all we have in solution is the weak base.

01:30 To carry out the ph of a weak base solution, we simply need to first solve for the hydroxide concentration.

01:40 The hydroxide concentration, as you recall, for a weak base is simply equal to the square root of kb 5 .2 times 10 to negative 4, multiplied by the concentration of the weak base .1 molar.

01:56 This is our hydroxide concentration.

01:58 If we divide this into kw, we will get the hydronium concentration.

02:04 If we take the negative log of the hydrogenium concentration, we will get our ph of 11 .86.

02:12 Next, we have added 10 milliliters of the strong acid to our weak base.

02:21 Because we started with 20 milliliters at 0 .1 molar of the weak base, and we are adding 0 .1 molar hcl, then we have a 20 -millimeter equivalence point.

02:37 So if we are at 10 -milliliter, we are halfway to the equivalence point, and as you recall, the ph at the equivalence point equals p -k -a.

02:47 P -k -a is simply kb divided into k -w.

02:51 This will give us our k -a value.

02:54 We take the negative log of that.

02:56 We'll get our p -k -a, and we'll recognize that at half equivalence, ph equals p -k -a.

03:02 If you didn't recognize this, you could still use the henderson -hasselbalch equation, and you will get the same answer.

03:09 Ph will be equal to p -k -a plus the log of the moles of base.

03:15 The moles of base will be equal to the moles of base we start with, which is 20 -mill liters or 0 .02 liters at 0 .1 molar, minus the moles of strong acid added, which will be the 10 -mill liters, or 0 .01, liters multiplied by its molarity of 0 .1.

03:34 This will be the moles of strong base left in solution after 10 milliliters, i'm sorry weak base left in solution after 10 milliliters of strong acid has been added.

03:44 We divide that by the moles of weak acid formed.

03:48 The moles of weak acid formed will be equal to the moles of strong acid added, which is our 10 milliliters, 0 .01 liters at 0 .1 molar.

04:00 This also gives us our ph equal to pca at 10 .72.

04:06 The next calculation c is similar to the previous.

04:11 We still have a buffer solution.

04:13 We are still pre -equivalence at 15 milliliters, where equivalence is 20 milliliters, so we still use the henderson -hasselbalch equation.

04:21 Ph is equal to pca plus the log of moles of base, which will be the moles of weak base we start with, again 20 millilitres at 0 .1 molar, minus the 1.

04:31 The moles of strong acid added, which can be calculated by taking the volume of strong acid added 15 milliliters or 0 .015 liters multiplied by the molarity.

04:41 We then divide that by the moles of weak acid formed, which will be equal to the moles of strong acid added, the same calculation as we did up here...

Please give Ace some feedback