A solution consisting of 25.00 g NH4 Cl in 178 mL of water is titrated with 0.114 M KOH. (a) H

A solution consisting of 25.00 g NH4 Cl in 178 mL of water...

A solution consisting of $25.00 \mathrm{~g} \mathrm{NH}_{4} \mathrm{Cl}$ in $178 \mathrm{~mL}$ of water is titrated with $0.114 \mathrm{M} \mathrm{KOH}$. (a) How many milliliters of $\mathrm{KOH}$ are required to reach the equivalence point? (b) Calculate $\left[\mathrm{Cl}^{-}\right],\left[\mathrm{K}^{+}\right],\left[\mathrm{NH}_{3}\right],$ and $\left[\mathrm{OH}^{-}\right]$ at the equivalence point. (Assume that volumes are additive.) (c) What is the $\mathrm{pH}$ at the equivalence point?

A solution consisting of $25.00 \mathrm{~g} \mathrm{NH}_{4} \mathrm{Cl}$ in $178 \mathrm{~mL}$ of water is titrated with $0.114 \mathrm{M} \mathrm{KOH}$.
(a) How many milliliters of $\mathrm{KOH}$ are required to reach the equivalence point?
(b) Calculate $\left[\mathrm{Cl}^{-}\right],\left[\mathrm{K}^{+}\right],\left[\mathrm{NH}_{3}\right],$ and $\left[\mathrm{OH}^{-}\right]$ at the
equivalence point. (Assume that volumes are additive.)
(c) What is the $\mathrm{pH}$ at the equivalence point?

Key Concepts

Equivalence Point

The equivalence point in a titration is the stage at which the number of moles of titrant added exactly equals the number of moles of the substance being titrated, based on the stoichiometric ratios defined in the balanced chemical equation. At this point, the reactant has been completely neutralized, although the pH of the solution may not necessarily be neutral due to the nature of the resulting species.

pH Calculation

pH calculation involves determining the concentration of hydrogen or hydroxide ions in a solution to assess its acidity or basicity. In titration, after reaching the equivalence point, the pH is often determined by the equilibrium behavior of the resulting species, such as through hydrolysis reactions, and calculating it requires an understanding of equilibrium constants and the relationship between ion concentrations and pH.

Hydrolysis

Hydrolysis in acid-base titrations refers to the reaction of the ions formed (especially those from weak acids or bases) with water, which can cause the pH of the solution to shift away from neutral. For example, the conjugate base formed from a weak acid may undergo hydrolysis to produce hydroxide ions, influencing the pH at the equivalence point.

Molarity and Volume Relationships

Molarity is a measure of concentration defined as the number of moles of a solute per liter of solution. In titration problems, understanding the relation between molarity, volume, and moles is essential, as it allows the conversion between measured volumes and the number of moles participating in a chemical reaction.

Titration

Titration is an analytical technique used to determine the concentration of an unknown solution by gradually adding a solution of known concentration (the titrant) until the chemical reaction between the two solutions reaches completion. This process relies on precise measurement of volumes and an understanding of the reaction stoichiometry between the analyte and titrant.

Acid-Base Neutralization

Acid-base neutralization involves the reaction between an acid and a base to form a salt and water. In the context of titration, the reaction is used to determine the concentration of the analyte by matching it with a base or acid of known concentration. Understanding the neutralization reaction is crucial to calculating the moles of reactants and products and for analyzing the chemical species present at various stages of the titration.

Stoichiometry

Stoichiometry is the quantitative relationship between reactants and products in a chemical reaction. It involves using the coefficients from the balanced chemical equation to calculate the number of moles involved, which is fundamental in titration for determining how much titrant is required to reach the equivalence point or to predict the concentrations of molecules present after the reaction.

Transcript

00:01 So our reaction is ammonium ion, reacting with our strong base, oh minus, to give us nh3 and h2o.

00:15 So we're starting with 25 grams of our ammonium chloride.

00:21 That's where we're getting our ammonium ion.

00:24 So let's go ahead and change grams to moles.

00:27 And the molar mass is 53 .50.

00:30 So that is 0 .467 moles of our ammonium ion.

00:39 So we know that we'll also react 0 .467 moles of our oh minus.

00:48 And we know we're also make 0 .467 moles of our nh3.

00:57 So we can take the moles of hydroxide and divide it by the molarity, which is 0 .114 molar.

01:04 And that will give us the volume, which is 4 .10 liters.

01:14 So let's find the concentration of our chloride ion.

01:19 As we said, we had 0 .467 moles of ammonium chloride.

01:26 And let's divide that by the total volume, which is going to be our 4 .10 liters plus our original 178 milliliters, or 0 .178 liters.

01:42 So that will give us a concentration of 0 .109 molar ammonium chloride or chloride.

01:53 Similarly, we started with 0 .114 molar k -oh, but the k in there is going to be a spectator ion, so it's not going to change, but it's going to get diluted over time.

02:08 So we started with 0 .178 liters, and we end up with our total volume again.

02:19 Okay? so again, we end up with 0 .109 molar k plus...

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