Quel est le pH final de la solution obtenue en ajoutant...

Quel est le pH final de la solution obtenue en ajoutant successivement à $50 \mathrm{~mL}$ d'une solution $0,1 \mathrm{M}$ d'acide bromhydrique $\mathrm{HBr}$ : 1) $20 \mathrm{~mL}$ d'une solution de soude $0,2 \mathrm{M}$, 2) $200 \mathrm{~mL}$ d'eau pure, 3) $100 \mathrm{~mL}$ d'une solution $0,05 \mathrm{M}$ d'acide perchlorique $\mathrm{HClO}_4$, 4) $130 \mathrm{~mL}$ d'une solution $0,5 \mathrm{M}$ de nitrate de potassium $\mathrm{KNO}_3$, 5) $1 \mathrm{~g}$ de potasse $\mathrm{KOH}$ solide.

Quel est le pH final de la solution obtenue en ajoutant successivement à $50 \mathrm{~mL}$ d'une solution $0,1 \mathrm{M}$ d'acide bromhydrique $\mathrm{HBr}$ :
1) $20 \mathrm{~mL}$ d'une solution de soude $0,2 \mathrm{M}$,
2) $200 \mathrm{~mL}$ d'eau pure,
3) $100 \mathrm{~mL}$ d'une solution $0,05 \mathrm{M}$ d'acide perchlorique $\mathrm{HClO}_4$,
4) $130 \mathrm{~mL}$ d'une solution $0,5 \mathrm{M}$ de nitrate de potassium $\mathrm{KNO}_3$,
5) $1 \mathrm{~g}$ de potasse $\mathrm{KOH}$ solide.

Key Concepts

Acid?Base Reaction Stoichiometry

This concept involves calculating the moles of acid and base using their volumes and concentrations, and determining how they neutralize each other. In acid-base titrations, the stoichiometry of the reaction is essential to establish which reactant is in excess after mixing, which in turn influences the pH of the final solution.

Neutralization Reactions of Strong Acids and Bases

Strong acids and strong bases react completely with each other in a neutralization reaction, meaning that their reaction goes to completion under typical conditions. Recognizing this complete reaction allows one to determine the amount of acid or base remaining after the reaction, which is a primary factor in determining the resulting pH.

Dilution Effects on Concentration

When additional water or solutions are added to a reacting solution, the concentrations of the species present are reduced. Understanding dilution is vital because the pH depends on the concentration of hydrogen or hydroxide ions, and dilution can shift these concentrations, thereby affecting the overall pH of the solution.

Impact of Inert Salts on pH

Some salts, when added to a solution, do not participate in acid-base reactions and are considered inert with respect to pH changes. Recognizing when a salt is inert helps in understanding that such an addition primarily affects ionic strength or solution volume without altering the pH directly.

Overall pH Calculation

The final pH of a solution is determined by the net effect of all contributions—from acid-base neutralization, leftover acid or base, and dilution factors. Calculating pH systematically involves accounting for the moles of reactive species, the total volume after all additions, and applying the definition of pH, which is the negative logarithm of the hydrogen ion concentration.

Transcript

00:01 Consider the titration of a 35 -millimeter sample of 0 .175 molar hbr with 0 .2 molar k -o -h.

00:07 Determine each quantity, a, the initial ph.

00:11 So initially, that's before you add any of our base.

00:13 So we only have that strong acid, hbr, present.

00:17 And because it's strong, it completely dissociates.

00:19 And so our concentration of h -plus is equal to the concentration of the hbr.

00:25 So i can take that molarity and plug that right into our ph equation.

00:29 So ph equals the negative log of .175, and so the ph when you calculate that is 0 .76.

00:39 Now for the solution, we want to find the volume of added base required to reach the equivalence point.

00:44 Because there's one hydrogen and one hydroxide, we can use this equation, and we can plug in the molarity and the volume of the analyte, and then we can plug in the molarity of the titrant and solve for the volume.

01:05 When you solve for the volume, you get 30 .6 milliliters of that k -oh.

01:13 Now we need to calculate the ph at 10 milliliters of added base.

01:17 To do this, we first need to find the moles of the hbr.

01:21 So we're going to convert the milliliters to liters, and then multiply by the molarity of the hbr, and that will give us the moles.

01:31 So we have 0 .06, 1, 2, 5 moles.

01:36 And because it completely dissociate it is equal to our moles of hydrogen ions.

01:45 Now let's go ahead and find the moles of the base that we added.

01:48 10 millilators change that to liters and multiply by the molarity of the base.

01:59 And you get 0 .002 moles of our hydroxide, again because it's strong, it completely dissociates.

02:08 So i have more hydrogen ions than i have hydroxide ions...

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