A 0.4000 M solution of nitric acid is used to titrate 50.00 mL of 0.237 M barium hydroxide. (As

A 0.4000 M solution of nitric acid is used to titrate 50.00...

Key Concepts

pH Calculation in Strong Acid/Strong Base Reactions

In titrations involving strong acids and strong bases, the pH is determined by the complete dissociation of these substances in water. Calculations of pH before any titrant is added, at intermediate points, and at the equivalence point are based on the concentrations of H? or OH? ions, using formulas such as pH = -log[H?] or pOH = -log[OH?] combined with the relation pH + pOH = 14.

Relationship of Molarity, Volume, and Moles

A fundamental concept in titrations is the relationship between molarity (concentration), volume, and moles, described by the equation moles = molarity × volume. This relationship is used to convert volume measurements into moles of reactants, which, when applied along with stoichiometric ratios from the balanced chemical equation, allows for precise calculations during the titration process.

Titration Stoichiometry

Titration stoichiometry involves using the balanced chemical equation to relate the moles of titrant to the moles of the substance being titrated. This stoichiometric relationship allows the calculation of the volume of titrant required to reach the equivalence point by equating the appropriate mole ratios from the balanced chemical equation.

Half-Equivalence Point in Titration

The half-equivalence point occurs when half of the analyte has been neutralized. Although buffer behavior is most commonly discussed for weak acid or weak base systems, in strong acid–strong base titrations, this point is used to illustrate the progression of the titration curve. At the half-equivalence point, the concentration of the remaining base (or acid) is exactly half of its initial value, which is reflective in the corresponding pH value, providing insight into the titration kinetics and progression.

Net Ionic Equation

The net ionic equation simplifies all reactions by only showing the species that undergo change, excluding spectator ions. In strong acid–strong base titrations, it generally reduces to the reaction between hydrogen ions (H?) and hydroxide ions (OH?) to form water, making it easier to see the stoichiometry of the neutralization reaction.

Acid-Base Titration

Acid-base titration is an analytical technique used to determine the concentration of an acid or base by reacting it with a titrant of known concentration. In such titrations, the reaction typically involves a neutralization process where the acid reacts with the base to form water and a salt, and volumes and concentration measures are used to reach the equivalence point.

Equivalence Point

The equivalence point in a titration is reached when the amount (in moles) of titrant added exactly neutralizes the analyte present in the solution. At this point, all of the reactant from one solution has reacted completely with the titrant, and the composition of the solution depends on the characteristics of the resultant species and the dilution from the added titrant.

Transcript

00:01 Okay, so we're reacting a strong acid, which is nitric acid, with a strong base, which is barium hydroxide.

00:08 So the first thing we're going to do is write the net ionic equation.

00:12 So the net ionic equation for strong acid and a strong base is always the same.

00:17 So h plus oh minus gives us water.

00:21 The things that will be present at the equivalence point will be the barium ion that came from the barium hydroxide.

00:28 And the nitrate ion that came from the nitric acid and of course our water.

00:39 Since we're going to do some stoichiometry here, i'm going to write out the complete reaction rather than net ionic equation.

00:46 It helps us see that there's a two to one ratio here between the barium hydroxide and the nitric acid.

00:55 I find it useful to go ahead and write that out rather than using a net ionic equation.

01:05 Well, they've given us some information about our barium hydroxide.

01:09 Okay, so i've got the molarity is 0 .237 molar.

01:14 So if i multiply by its volume and liters, i'll get the moles of barium hydroxide.

01:22 1 .19 moles.

01:29 So if i want to then find out how much nitric acid would react with that, i'll just do our mole ratio.

01:37 Right, moles of barium hydroxide to nitric acid, and that's 2 to 1.

01:46 So we'll see that we needed or 0 .237 moles of nitric acid will have reacted.

01:58 So i'm going to go ahead and take those moles, 027 moles.

02:06 And i have its molarity.

02:07 So i'm going to divide by its molarity, which is moles per liter, so that'll give us the liters that we needed to add.

02:15 So 0 .0593 liters or 593 milliliters, or 59 .3 milliliters.

02:24 Of our nitric acid was added at the equivalence point.

02:32 So the ph before anything is added, all we have is 0 .237 molar barium hydroxide.

02:40 We haven't added any nitric acid yet.

02:44 So i'm going to multiply that by two to get the molarity of my hydroxide ion, my oh minus.

02:51 So 0 .474 molar, right? because each barium hydroxide has two oh h minuses in it...

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