L'addition d'acide chlorhydrique dans une solution de nitrate de plomb provoque la précipitation

L'addition d'acide chlorhydrique dans une solution de...

Key Concepts

Stoichiometry

Stoichiometry concerns the quantitative relationships between reactants and products in a chemical reaction. It is essential for determining how much of one substance is required to react with a given amount of another substance based on their mole ratios in the balanced chemical equation.

Balanced Chemical Equation

A balanced chemical equation represents the reactants and products in a chemical reaction with their respective coefficients adjusted so that the number of atoms for each element is conserved. This ensures that the law of conservation of mass is obeyed and allows for accurate stoichiometric calculations.

Precipitation Reaction

A precipitation reaction involves the formation of an insoluble solid (the precipitate) when two soluble salts in aqueous solution react. Understanding which combinations of ions will result in an insoluble product is key to predicting and controlling such reactions.

Molarity and Volume Relationships

Molarity, defined as the number of moles of solute per liter of solution, is a central concept in quantitative solution chemistry. It is used to relate the concentration of a solution to the volume, allowing the calculation of the number of moles present in any given volume, which is crucial for determining the amounts needed for complete reactions.

Transcript

00:01 So we're going to combine our two solutions, so they're going to become diluted.

00:06 So if we start with our equation for dilutions, m1v1 equals m2 v2, our new molarity of each solution will be its original molarity, times its volume, divided by the new volume.

00:22 And this new volume is the total volume.

00:25 So the total volume of the two solutions added together, which will be 38 milliliter.

00:31 Okay, so let's find out our concentration of our solutions once we combine them.

00:39 So the concentration of that mercury 1 is the 0 .0021 molar times its 13 milliliters over the total milliliters.

00:56 So it's going to give us 7 .18 times 10 to the negative 4 molar.

01:02 And the concentration of our chloride ion is 0 .15.

01:10 Molar times the 25 milliliters that we started with of that again divided by the total will give its diluted new molarity and our equation is our solid that's the precipitate we're going to form in equilibrium with its ions 2 2 plus plus 2cl minus so our expression for q is the same as k and so mercury 2 2 plus concentration times a concentration of chloride squared.

01:56 So we'll go ahead and plug all that in.

02:01 So that's going to give us 7 .18 times 10 of the negative 4 times 9 .87 times 10 of the negative 3 squared.

02:17 So q is going to come out to be 7 .1 times 10.

02:22 Of the negative 6, and this is larger than 1 times 10 of the negative 18, which is our value for ksp.

02:32 So since q is larger, we will say yes, so precipitate forms.

02:43 Okay, so to find all these concentrations that we're looking for at equilibrium, we're going to set up an ice box.

02:50 We'll ignore our solid, and we'll write down our starting concentrations, 7 .18 times 10 to the negative 4 and 9 .87 times 10 of the negative 3.

03:05 This will be minus x and minus 2x.

03:09 So 7 .18 times 10 of the negative 4 minus x and 9 .87 times 10 of the minus 3 minus 2x.

03:22 Okay, so our expression, so is our ksp is 1 times 10 of the negative 18.

03:30 So this is going to be 7 .18 times 10 of the negative 4 minus x times 9 .87 times 10 of the negative 3 minus 2x, and we need to square that.

03:49 So if you take a look at that equation, that's not going to be much fun to solve at all...

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