A solution of Ba(OH) 2 has a K sp of 5.0 x 10 − 3 . i) Determine the pH of this solution. ii) Determine the pH if Ba(OH) 2 was added to a solution containing 3.2 M of BaF 2 and 0.94 M of Al(OH) 3 .

Here we're told that a solution of barium hydroxide has a Ksp of 5.0 times 10 to the negative 3. For this first part, we have to determine the pH of the solution. So for part 1, we have Ksp of a compound, which means it is a solid, and we show how it breaks up into ions. It breaks up into 1 barium ion plus 2 hydroxide ions. With KSP, we're dealing with an ICE chart, and we ignore solids and liquids. Here these products initially are 0 when we're making them, so this is plus x, plus 2 x, plus x, plus 2 x. Ksp just equals products, so Ksp equals barium ion times hydroxide ion. And again it's going to be squared because of the coefficient of 2. So now we have Ksp is 5.0 times 10 to the minus 3 equals x times 2 x squared. This will equal x times 4 x squared, which equals 4x cubed, which is still equal to 5.0 times 10 to the minus 3. Here we need to isolate x, so divide both sides by 4. Then we're going to take the cube root. So x cubed equals 1.25 times 10 to the minus 3. Take the cube root of both sides, x equals 0.1077. This is important because we need to find pH. We're going to say Oh minus concentration at equilibrium is equal to 2 x. So take this x value that we just found and plug it in. So it's going to be 0.2154. Taking the negative log of that will give me pOH because pOH equals the negative log of Oh- concentration. And if I know my pOH, I know pH because pH equals 14 minuteus pOH. So that comes out to 13.33. For part 2, they're saying determine the pH of barium hydroxide is added to a solution containing 3.2 molar of barium fluoride and 0.94 molar of aluminum hydroxide. So for this question here, we're going to say that the pH of a base like barium hydroxide is dependent on the concentration of your Oh minus, like we saw in part 1. So, the key to this question is realizing we have 0.94 molar of aluminum hydroxide. There are 3 hydroxide ions involved, so when I multiply this value by 3 I get 2.82 molar of hydroxide ion concentration. Taking the negative log of that value gives me my pOH. And since it's a value greater than 1, it's going to be a pretty low pOH, it's going to be negative 0.45. PH equals 14 minuteus pOH, so it equals 14 minuteus a minus 0.45. Here, our pH is off scale, greater than 14. Remember we said in earlier videos, the pH scale we're used to seeing it between 0 14. That's only true if the concentration of our strong acid or base has a molarity that is less than 1. Here, the molarity of our base was a value much greater than 1. So that's gonna mean that our pH is gonna be off scale. It's gonna be, lower than 0 or it could be greater than 14. Here, since the strong bass is pretty concentrated, it has a pH that's greater than 14.

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18. Aqueous Equilibrium

Ksp: Common Ion Effect

General Chemistry

18. Aqueous Equilibrium

Ksp: Common Ion Effect

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Multiple Choice

A solution of Ba(OH)₂ has a K_sp of 5.0 x 10⁻³.
i) Determine the pH of this solution.
ii) Determine the pH if Ba(OH)₂ was added to a solution containing 3.2 M of BaF₂ and 0.94 M of Al(OH)₃.

i. 2.30
ii. 14.03

i. 0.67
ii. 13.55

i. 13.33
ii. 14.45

i. 2.30
ii. 13.65

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Verified step by step guidance

Step 1: Understand the dissociation of Ba(OH)2 in water. Ba(OH)2 dissociates into Ba²⁺ and 2 OH⁻ ions. The balanced equation is: Ba(OH)₂(s) ⇌ Ba²⁺(aq) + 2 OH⁻(aq).

Step 2: Use the solubility product constant (Ksp) to find the concentration of OH⁻ ions. The expression for Ksp is: Ksp = [Ba²⁺][OH⁻]². Given Ksp = 5.0 x 10⁻³, set up the equation: 5.0 x 10⁻³ = (s)(2s)², where s is the solubility of Ba(OH)₂.

Step 3: Solve for s, the solubility of Ba(OH)₂, which represents the concentration of Ba²⁺ ions. Then, calculate the concentration of OH⁻ ions as 2s.

Step 4: Calculate the pOH of the solution using the formula: pOH = -log[OH⁻].

Step 5: Convert pOH to pH using the relationship: pH + pOH = 14. For the second part, consider the common ion effect and the initial concentrations of BaF₂ and Al(OH)₃, and repeat the process to find the new pH.