Pyridine, C5 H5 N, has Kb of 1.7 ×10^-9 . A 38.00-mL sample of pyridine requires 29.20 mL of 0

Pyridine, C5 H5 N, has Kb of 1.7 ×10^-9 . A 38.00-mL sample...

Pyridine, $\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{~N},$ has $K_{\mathrm{b}}$ of $1.7 \times 10^{-9} .$ A $38.00-\mathrm{mL}$ sample of pyridine requires $29.20 \mathrm{~mL}$ of $0.332 \mathrm{M}$ HBr to reach the equivalence point. (a) Write a balanced net ionic equation for the titration. (b) How many grams of pyridine are in the sample? (c) Calculate $\left[\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{~N}\right],\left[\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{NH}^{+}\right],\left[\mathrm{H}^{+}\right]$ and $\left[\mathrm{Br}^{-}\right]$ at the equivalence point. (Assume that volumes are additive.) (d) What is the $\mathrm{pH}$ at the equivalence point?

Pyridine, $\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{~N},$ has $K_{\mathrm{b}}$ of $1.7 \times 10^{-9} .$ A $38.00-\mathrm{mL}$ sample of pyridine requires $29.20 \mathrm{~mL}$ of $0.332 \mathrm{M}$ HBr to reach the equivalence point.
(a) Write a balanced net ionic equation for the titration.
(b) How many grams of pyridine are in the sample?
(c) Calculate $\left[\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{~N}\right],\left[\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{NH}^{+}\right],\left[\mathrm{H}^{+}\right]$ and $\left[\mathrm{Br}^{-}\right]$ at the equivalence
point. (Assume that volumes are additive.)
(d) What is the $\mathrm{pH}$ at the equivalence point?

Key Concepts

Conjugate Acid Hydrolysis

At the equivalence point of a weak base-strong acid titration, the resulting solution predominantly contains the conjugate acid of the weak base. This conjugate acid undergoes hydrolysis, reacting with water to produce hydronium ions. Understanding this process is important because it determines the solution’s pH and provides insight into the acidic nature of the conjugate species formed.

Equilibrium Constant Relationships

The equilibrium constant relationships, particularly the linkage between a weak base's Kb and its conjugate acid's Ka via the water ionization constant (Kw), are crucial in determining the behavior of the conjugate acid in solution. This conversion is essential when calculating the pH at the equivalence point, as it dictates the extent of hydrolysis of the conjugate acid formed during titration.

pH Calculation

pH calculation in the context of acid-base titrations, particularly at the equivalence point, involves determining the concentration of hydronium ions produced by the hydrolysis of the conjugate acid. This requires setting up an equilibrium expression based on the relevant Ka, solving for the hydronium ion concentration, and finally converting that concentration into a pH value. This concept is central to interpreting titration curves and understanding the acidity of the solution after neutralization.

Net Ionic Equation

The net ionic equation represents the essence of a chemical reaction by including only the reacting species and omitting the spectator ions. In titrations, the net ionic equation simplifies the reaction to its fundamental acid-base neutralization process and is crucial for understanding reaction mechanics and subsequent equilibrium analyses.

Acid?Base Titration

Acid?base titration involves the gradual addition of a titrant (in this case, a strong acid) to a solution containing an analyte (a weak base) until reaching the equivalence point, where the number of moles of acid added equals the number of moles of base initially present. This concept is critical because it allows for the quantitative determination of the concentration of the unknown solution through stoichiometric relationships.

Reaction Stoichiometry

Reaction stoichiometry applies the mole concept to determine the relationships between reactants and products in a chemical reaction. In titration problems, it is used to convert volumes and concentrations to moles, allowing the calculation of unknown quantities such as the mass of the analyte by equating the moles of titrant delivered to the analyte present at the equivalence point.

Transcript

00:01 The balanced equation here is going to be our weak base, our mean, reacting with a strong acid, so h plus, and giving us the conjugate acid of our mean.

00:18 Okay, so there's our balanced equation.

00:21 And we're going to try to figure how many grams of periodine we have.

00:25 So i started with 0 .332 molar and 29 .2 milliliters, so 0 .0292 milliliters, so 0 .0292 .2.

00:34 Liters of our strong acid.

00:38 So that gives us 0 .009 .69 moles of age plus.

00:45 So there must have been that same amount of our perioding, our weak base present.

00:55 So i'll just go ahead and multiply that by its molar and mass grams per mole.

01:01 And that will give us 0 .767 grams of our weak base to start with.

01:09 I'm going to do c and d together here.

01:12 All right.

01:13 So to start with, we want to get a concentration of our product, okay, our conjugate acid.

01:22 So we know that we're going to produce that same number of moles because everything's one to one.

01:29 And then to find its concentration, i need the total volume.

01:33 Well, i had 38 milliliters plus 29 .2 millimeters, which gives me 67 .2 millimeters.

01:43 So the total volume is going to be 0 .0672 liters.

01:51 So that's going to give us 0 .144 molar.

01:56 So if we want to know, let's go ahead and set up our ice box.

02:04 So our, because this is what we're forming at the equivalence point.

02:09 And we want to know it's ph and some stuff about it...

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