A 50.0-mL sample of NaHSO3 is titrated with 22.94 mL of 0.238 M KOH. (a) Write a balanced net i

step 1 Okay, so we're going to react a weak acid with a strong base. Okay, so the hydrogen sulfite, whi

A 50.0-mL sample of NaHSO3 is titrated with 22.94 mL of...

A $50.0-\mathrm{mL}$ sample of $\mathrm{NaHSO}_{3}$ is titrated with $22.94 \mathrm{~mL}$ of $0.238 \mathrm{M} \mathrm{KOH}$. (a) Write a balanced net ionic equation for the reaction. (b) What is $\left[\mathrm{HSO}_{3}^{-}\right]$ before the titration? (c) Find $\left[\mathrm{HSO}_{3}^{-}\right],\left[\mathrm{SO}_{3}^{2-}\right],\left[\mathrm{OH}^{-}\right],\left[\mathrm{K}^{+}\right],$ and $\left[\mathrm{Na}^{+}\right]$ at the equivalence point. (d) What is the $\mathrm{pH}$ at the equivalence point?

A $50.0-\mathrm{mL}$ sample of $\mathrm{NaHSO}_{3}$ is titrated with $22.94 \mathrm{~mL}$ of $0.238 \mathrm{M} \mathrm{KOH}$.
(a) Write a balanced net ionic equation for the reaction.
(b) What is $\left[\mathrm{HSO}_{3}^{-}\right]$ before the titration?
(c) Find $\left[\mathrm{HSO}_{3}^{-}\right],\left[\mathrm{SO}_{3}^{2-}\right],\left[\mathrm{OH}^{-}\right],\left[\mathrm{K}^{+}\right],$ and $\left[\mathrm{Na}^{+}\right]$ at
the equivalence point.
(d) What is the $\mathrm{pH}$ at the equivalence point?

Key Concepts

Hydrolysis and pH Calculation at Equivalence

At the equivalence point, the reaction products, such as the salt formed from a weak acid or base, can undergo hydrolysis by reacting with water. This hydrolysis can alter the pH of the solution, making it basic or acidic. Understanding and calculating the pH at the equivalence point thus requires an analysis of the equilibrium between the hydrolyzed species and water using the appropriate equilibrium constants.

Concentration Calculations

Concentration calculations in the context of titrations involve determining the molarity of the analyte and its reaction products by using the volume and molarity of the titrant added and the initial volume of the analyte solution. These calculations are essential for establishing both the initial conditions of the titration and the concentrations of species present after the reaction, including any spectator ions.

Equivalence Point

The equivalence point in a titration is reached when the number of moles of titrant added exactly equals the number of moles of the substance being titrated, as dictated by the reaction stoichiometry. At this point, the original reactant is completely converted into product(s), and the properties of the resulting solution (such as pH) may be affected by the chemical nature of the products.

Acid-Base Titration

Acid-base titration is a method used to determine the concentration of an acid or base in a solution by gradually adding a reactant of known concentration (the titrant) until the reaction reaches its equivalence point. This process relies on the neutralization reaction between the acid and base, enabling quantitative analysis of the analyte based on the stoichiometry of the reaction.

Net Ionic Equation

A net ionic equation summarizes only the chemical species that actively participate in a reaction, omitting spectator ions that do not change during the process. In acid-base reactions, the net ionic equation highlights the transfer of protons, clarifying the actual chemical change occurring as an acid or a base reacts with its counterpart.

Reaction Stoichiometry

Reaction stoichiometry involves using a balanced chemical equation to relate the quantities of reactants and products in a chemical reaction. In titration, stoichiometry is key for determining how many moles of titrant are required to completely react with a given number of moles of the analyte, thereby allowing calculations of unknown concentrations.

Transcript

00:01 Okay, so we're going to react a weak acid with a strong base.

00:04 Okay, so the hydrogen sulfite, which comes from sodium hydrogen sulfite, so h.

00:10 So h .o3 minus, that's going to end up being our weak acid.

00:14 Okay, this came from the nahs.

00:18 So3 is going to react with oh minus.

00:24 That came from our k -o -h.

00:27 And, of course, they'll make water, and the conjugate base sulfite.

00:36 So that's our net ionic equation.

00:39 So now we want to figure out what's the concentration of the hydrogen sulfite before we do any titrating.

00:46 Okay.

00:49 Well, we can find the moles of oh minus, or k -o -h, because they've given us the molarity.

00:56 0 .238 molar.

01:00 And the volume, we're going to change that to liters to 2 -94 liters.

01:05 So we know that 0 .00546 moles of k -o -h or oh -minas reacted.

01:17 Therefore, we must have had the same number of moles.

01:20 0 .0 -546 moles of our hydrogen sulfite.

01:29 And since we added 0 .050 liters, that molarity must be 0 .109 mole.

01:40 So that was the initial molarity of the hs.

01:45 H .s .o .3 minus.

01:48 So now we're going to try to figure out a bunch of concentrations at the equivalence point.

01:56 Okay, so what's happening at the equivalence point? well, at the equivalence point, i think we can see that we have s .o .3 .2 negative, which is a weak base.

02:09 So we're going to want to write the equation for that weak base ionizing.

02:14 So we'll write the hydrolysis reaction.

02:18 So that's going to react with water to make h .s .o .3 minus and some hydroxide.

02:30 So if we set up an ice box to keep track of what's happening, we need the initial concentration of the s .o .3 .2.

02:37 Well, we know that if we've reacted this many moles of this and this many, moles of this, we must have formed the same number of moles of the sulfite.

02:57 So now to get the concentration of that, we need to dissolve, we need to divide by the total volume.

03:04 All right.

03:05 Well, the total volume, if we look, we had 22 .94 millimeters.

03:15 And we had 50 millimeters.

03:19 So our total volume is going to be 72 .9.

03:22 94 milliliters.

03:25 So i'll come over here and divide by that, but in liters.

03:28 0 .7294 liters...