Calculate the solubility of Ag3 PO4 in a solution buffered...

Key Concepts

Solubility Equilibrium

This concept describes the dynamic balance between a solid ionic compound and the ions it forms when dissolved in a solution. In the case of sparingly soluble salts, such as many metal phosphates, the equilibrium establishes a constant value (the solubility product) which is a measure of the extent to which the compound can dissolve. It involves the dissolution reaction reaching a state where the rate of dissolution equals the rate of precipitation.

Solubility Product (Ksp)

The solubility product constant, Ksp, is a key parameter in determining the solubility of a sparingly soluble salt. It quantitatively expresses the product of the concentrations of the constituent ions, each raised to the power of their stoichiometric coefficients, at equilibrium. This constant is essential to calculate the amount of solute that can dissolve under given conditions.

Buffer and pH Effect

The pH of a solution can strongly influence the solubility of compounds, particularly those that produce or consume H+ ions upon dissolution. In buffered solutions, the pH is held constant, which can shift the solubility equilibrium, especially when the dissolution process involves ions that form acid-base equilibria. Understanding the effect of pH is crucial for predicting how a change in acidity or basicity alters the solubility of a compound.

Polyprotic Acid Speciation

Polyprotic acids, such as phosphoric acid, can donate more than one proton and exist in several protonated forms depending on the pH of the solution. The distribution among various species (for example, PO4^3-, HPO4^2-, and H2PO4^-) depends on the pKa values of the acid. This speciation affects the solubility equilibria of salts containing polyprotic anions because only specific protonated forms may participate in the dissolution equilibrium, thereby altering the effective concentration of the ion involved in the solubility product.

Transcript

00:02 Okay, so this video is going to go for problem 20 in chemistry second edition, which is in chapter 15.

00:12 And this is equilibrium of other reaction classes, and this is specifically solubility and the common ion effect.

00:20 So what the common ion effect is, is if you have an ion in solution already, that is similar to, that is the same ion that you have, that your solid is.

00:33 Going to produce, it can prevent the dissolution of the solid.

00:42 So it can, it makes your calculations a little different and impacts your solubility and impacts how much you're going to make.

00:52 Okay, so this one is, what is the solubility of aluminum hydroxide and a buffer solution with a ph of 11? so if we look at this, a ph of 11, what does that mean? so that means we have a basic solution.

01:11 So we have a lot of oh minus in this buffer solution.

01:18 And so if we look at our solid, our solids also going to create oh minus.

01:23 So we already have hydroxide present in our solution before we start dissolving.

01:31 And so to do this, what we need to do is we need to set up a rice table or an ice table.

01:47 And this will help us figure out where to start.

02:06 Okay, so here we have our reaction.

02:08 And for an ice table, we need our initial.

02:14 We don't have to worry about solids because for equilibrium, i mean, they don't have a concentration.

02:19 So there's no way we can put them into a calculation.

02:23 But our initial concentration, we start out with zero, but we don't start out with zero hydroxide, right? we figured out that we have a basic solution.

02:34 We have some hydroxide in there.

02:36 And then, so we need to calculate the amount of hydroxide that we start out with.

02:44 So we have a ph 11 of 11.

02:50 And if we think about our ph equations, so our ph plus our poh is going to equal 14, right? so we can calculate our poh from that ph.

03:04 And that will be just 14 minus ph, which is 14 minus 11.

03:11 So our poh is 3.

03:18 And then to calculate our concentration of oh from poh, we can remember that p -o -h simply means the negative log of hydroxide.

03:33 So similar to our ph equals the negative log of h -plus, a concentration of h -plus.

03:40 If we take the inverse of this equation, we get the concentration of o -h is equal to 10 times negative p -o -h.

03:54 And we know our p -o -h is 3, so 10 to the negative 3, is our concentration of o -h.

04:01 Oh, h.

04:01 So we can go back up here, and we can put that in our place for our initial.

04:10 Now, our change, we're going to do the same thing.

04:12 So we're going to increase by some amount x, and then we're going to increase by some amount 3x, but we start with 10 to the negative third.

04:22 So it's going to be 10 to the negative third plus 3x...

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