# How to calculate ksp with temperature?

**Answer # 1 #**

To find Ksp, we first need a chemical equation describing calcium hydroxide dissolving:

Ca(OH)2→Ca2++2OH-

We also need the rate law for this equation:

Kc=2/

The numerator in this rate law, 2, is Ksp. So we need to figure out what and are. We can find them since we have the molarity (same thing as solubility) of the calcium hydroxide, 1.11*10-2 mol/L. We always assume the chemical equation occurs in a 1 L solution, which makes the calculations easier. So, we are starting with 1.11*10-2 moles of Ca(OH)2. By looking at the coefficients in the equation (if not written then the coefficient is 1), we see that for every mole of Ca(OH)2 put in, we get 1 mole of Ca2+ as a product. Therefore the moles of Ca2+ obtained is the number of moles of Ca(OH)2 put in, 1.11*10-2. Since this is still happening in a 1 L solution, =1.11*10-2 mol/L. As for OH-, the equation says that for every mole of Ca(OH)2 put in, we get 2 moles of OH-. Therefore the moles of OH- obtained is 2 times the number of moles of Ca(OH)2 put in, or 2.22*10-2, and like with , =2.22*10-2 mol/L.

Therefore Ksp=(1.11*10-2)(2.22*10-2)2=5.5*10-6 (do not plug in any measurement units).

**Answer # 2 #**

The solubility product constant is the equilibrium constant for the dissolution of a solid substance into an aqueous solution. It is denoted by the symbol Ksp.

Solubility is defined as a property of a substance called solute to get dissolved in a solvent in order to form a solution. The solubility of ionic compounds (which dissociate to form cations and anions) in water varies to a great deal. Some compounds are highly soluble and may even absorb moisture from the atmosphere, whereas others are highly insoluble.

Solubility depends on a number of parameters, amongst which the lattice enthalpy of salt and solvation enthalpy of ions in the solution are of most importance.

Salts are classified on the basis of their solubility in the following table.

Suppose barium sulphate along with its saturated aqueous solution is taken. The following equation represents the equilibrium set up between the undissolved solids and ions:

\(\begin{array}{l}BaSO_4 \overset{saturated ~solution ~in~ water}{\rightleftharpoons} Ba^{2+} (aq) + SO_4^{-} (aq)\end{array} \)

The equilibrium constant in the above case is:

In the case of pure solid substances, the concentration remains constant, and so we can say:

Here Ksp is known as the solubility product constant. This further tells us that solid barium sulphate when in equilibrium with its saturated solution, the product of concentrations of ions of both barium and sulphate is equal to the solubility product constant.

We now have a brief idea about the solubility of a compound and the factors affecting it. For any further query on this topic, call the mentor support team at BYJU’S.

**Answer # 3 #**

- Problem: Write the K
_{sp}expression for the following weak electrolytes: Mn(OH)_{3}(s), Sr_{3}(AsO_{4})_{2}(s), and Co_{2}S_{3}(s). - Calculating Solubility from K
_{sp}: - K
_{sp}= [Ca^{2}^{+}][F^{-}]^{2}= [2.14x10^{-}^{4}][4.28x10^{-}^{4}]^{2}= 3.9 x 10^{-}^{11}

**Answer # 4 #**

Now, imagine that we have the Ksp value for a specific compound. But this time, we want to determine the concentrations of ions present. Let's start with an easy one: calcium carbonate (CaCO3). Calcium carbonate has a Ksp of 8.7 * 10^-9 at 25 degrees Celsius.

We know that calcium carbonate breaks into calcium and carbonate ions, and we know that the equilibrium expression is Ksp = .

Calcium and carbonate ions dissociate from calcium carbonate in a 1:1 ratio, so we should have equal concentrations of each ion in solution. We can substitute x for the concentrations of each ion and plug in our known value of Ksp, then solve for x!

Ksp = (x)(x) = 8.7 * 10^-9

We find the value of x by taking the square root of 8.7 * 10^-9. x = 9.3 * 10^-5.

Since we set x equal to the concentrations of the ions, we know that the concentration of Ca^2+ and CO3^2- are each 9.3 * 10^-5 mol/L.

One more, and we're done! Let's find the concentrations of lead (II) chloride, PbCl2, at 25 degrees Celsius. The Ksp for this compound is 1.6 * 10^-5. That's a small number, so we'll expect the concentrations of our dissolved ions will be small amounts.

Like we've done before, let's write an equation showing how PbCl2 will break into ions (and reform). The law of mass action for this dissociation is Ksp = ^2.

We know that each PbCl2 compound breaks into 1 Pb^2+ ion and 2 Cl^- ions. This means that the concentration of Cl^- ions will be twice that of Pb^2+. If I set the concentration of Pb^2+ equal to x, then I must set the concentration of Cl^- to 2x. I can now plug these values into my equilibrium expression along with my value of Ksp.

Ksp = (x)(2x)^2 = 1.6 * 10^-5

Simplifying, I find that 4x^3 = 1.6 * 10^-5, and therefore x = 0.016.

This means that my concentration of Pb^2+ is 0.016 mol/L and the concentration of Cl^- is 0.032 mol/L. Those are tiny numbers, just like we thought.

Solubility is the ability to be dissolved in a given volume of a solvent at a given temperature. Solubility for a compound is expressed in terms of molarity (mol/L) and is temperature dependent. A substance is considered to be soluble when its solubility is greater than 1 g per 100 g of solvent. A substance is considered to be insoluble when its solubility is less than 0.1 g per 100 g solvent.

When a solution of a given volume has a maximum amount of a solid dissolved in it, it is said to be saturated. For solutions at equilibrium, the law of mass action is written only in terms of the ions produced. The equilibrium constant is written as Ksp and identified as the solubility product constant (or the solubility constant) for a solution at equilibrium.

Ksp can be calculated by writing the law of mass action for a solution and inputting the concentration of ions into the equation. This can be done using ion concentrations or based on solubility data. The concentrations of ions can be determined given only the Ksp value for a solution at equilibrium. In this case, write the law of mass action. Set the unknown concentrations equal to x. Make sure to account for stoichiometric ratios. Solve for x.

**Answer # 5 #**

The solubility product constant is the equilibrium constant for the dissolution of a solid substance into an aqueous solution. It is denoted by the symbol Ksp.

Solubility is defined as a property of a substance called solute to get dissolved in a solvent in order to form a solution. The solubility of ionic compounds (which dissociate to form cations and anions) in water varies to a great deal. Some compounds are highly soluble and may even absorb moisture from the atmosphere, whereas others are highly insoluble.

Solubility depends on a number of parameters, amongst which the lattice enthalpy of salt and solvation enthalpy of ions in the solution are of most importance.

Salts are classified on the basis of their solubility in the following table.

Suppose barium sulphate along with its saturated aqueous solution is taken. The following equation represents the equilibrium set up between the undissolved solids and ions:

\(\begin{array}{l}BaSO_4 \overset{saturated ~solution ~in~ water}{\rightleftharpoons} Ba^{2+} (aq) + SO_4^{-} (aq)\end{array} \)

The equilibrium constant in the above case is:

In the case of pure solid substances, the concentration remains constant, and so we can say:

Here Ksp is known as the solubility product constant. This further tells us that solid barium sulphate when in equilibrium with its saturated solution, the product of concentrations of ions of both barium and sulphate is equal to the solubility product constant.