Question 1

A 100.0 mL sample of 0.10 M Ca(OH)₂ is titrated with 0.10 M HBr.Determine the pH of the solution after the addition of 400.0 mL HBr.

Accepted Answer

The answer of A 100.0 mL sample of 0.10 M...

Question 2

Which of the following compounds solubility will not be affected by a low pH in solution?

Accepted Answer

AgCl is a soluble salt that is not affected by pH.

Question 3

A 100.0 mL sample of 0.10 M NH₃ is titrated with 0.10 M HNO₃.Determine the pH of the solution after the addition of 50.0 mL of HNO₃.The K_b of NH₃ is 1.8 × 10^-5.

Accepted Answer

The answer of A 100.0 mL sample of 0.10 M...

Question 4

A 100.0 mL sample of 0.10 M NH₃ is titrated with 0.10 M HNO₃.Determine the pH of the solution after the addition of 100.0 mL of HNO₃.The K_b of NH₃ is 1.8 × 10^-5.

Accepted Answer

The answer of A 100.0 mL sample of 0.10 M...

Question 5

A 100.0 mL sample of 0.18 M HClO₄ is titrated with 0.27 M LiOH.Determine the pH of the solution after the addition of 100.0 mL of LiOH.

Accepted Answer

1. Write the balanced chemical equation for the reaction between HClO and LiOH:

HClO + LiOH -> LiClO + H2O

2. Calculate the initial number of moles of HClO:

0.18 M HClO x 0.100 L = 0.018 mol HClO

3. Calculate the number of moles of LiOH added:

0.27 M LiOH x 0.100 L = 0.027 mol LiOH

4. Determine the limiting reagent:

HClO:LiOH = 0.018 mol:0.027 mol = 0.67:1

Since LiOH is in excess, it is the limiting reagent.

5. Calculate the number of moles of LiClO produced:

0.027 mol LiOH x 1 mol LiClO/1 mol LiOH = 0.027 mol LiClO

6. Calculate the concentration of LiClO:

0.027 mol LiClO/0.200 L = 0.135 M LiClO

7. Determine the pH of the solution:

To determine the pH, we need to calculate the concentration of H+ ions in the solution. Since LiOH is a strong base, it will react completely with HClO to form LiClO and H2O.

HClO + LiOH -> LiClO + H2O

Since HClO is a weak acid, the reaction can be approximated by:

HClO + OH- -> ClO- + H2O

Kb = Kw/Ka = 1.0 x 10^-14/3.0 x 10^-8 = 3.3 x 10^-7

Since [OH-] = [LiOH] = 0.027 mol/0.200 L = 0.135 M, we can use the Kb expression to calculate the concentration of H+:

Kb = [ClO-][H+]/[OH-] = 3.3 x 10^-7

[H+] = Kb[OH-]/[ClO-] = 3.3 x 10^-7 x 0.135 M/0.018 M = 2.5 x 10^-6 M

pH = -log[H+] = 12.65

Question 6

A 100.0 mL sample of 0.10 M Ca(OH)₂ is titrated with 0.10 M HBr.Determine the pH of the solution before the addition of any HBr.

Accepted Answer

The answer of A 100.0 mL sample of 0.10 M...

Question 7

A 100.0 mL sample of 0.20 M HF is titrated with 0.10 M KOH.Determine the pH of the solution after the addition of 300.0 mL of KOH.The K_a of HF is 3.5 × 10^-4.

Accepted Answer

After adding 300.0 mL of 0.10 M KOH to 100.0 mL of 0.20 M HF, the solution becomes basic as KOH is in excess. Calculate the moles of HF and KOH, determine the excess moles of KOH, and then find the pH using the concentration of OH⁻ ions. The pH is approximately 12.40.

Question 8

Which of the following compounds will have the highest molar solubility in pure water?

Accepted Answer

The answer of Which of the following compounds will have...

Question 9

A 100.0 mL sample of 0.10 M NH₃ is titrated with 0.10 M HNO₃.Determine the pH of the solution before the addition of any HNO₃.The K_b of NH₃ is 1.8 × 10^-5.

Accepted Answer

The pH of the solution before any HNO₃ is added is determined by the ammonia solution itself. Ammonia is a weak base, and its pH can be calculated using the formula for the pH of a weak base: pH = 14 - pOH, where pOH = -log(√(Kb × [NH₃])). Substituting the given values, pOH ≈ 2.87, so pH ≈ 14 - 2.87 = 11.13.

Question 10

A 100.0 mL sample of 0.10 M NH₃ is titrated with 0.10 M HNO₃.Determine the pH of the solution after the addition of 150.0 mL of HNO₃.The K_b of NH₃ is 1.8 × 10^-5.

Accepted Answer

The answer of A 100.0 mL sample of 0.10 M...

Question 11

A 100.0 mL sample of 0.20 M HF is titrated with 0.10 M KOH.Determine the pH of the solution after the addition of 400.0 mL of KOH.The K_a of HF is 3.5 × 10^-4.

Accepted Answer

The answer of A 100.0 mL sample of 0.20 M...

Question 12

A 100.0 mL sample of 0.20 M HF is titrated with 0.10 M KOH.Determine the pH of the solution before the addition of any KOH.The K_a of HF is 3.5 × 10^-4.

Accepted Answer

Since we are given the initial concentration of HF and its Ka, we can use the Henderson-Hasselbalch equation to find the initial pH of the solution:

pH = pKa + log([A-]/[HA])

where [A-] is the concentration of the conjugate base (F-) and [HA] is the concentration of the acid (HF).

pKa = -log(Ka) = -log(3.5 x 10^-4) = 3.46

[A-]/[HA] = 0/0.20 = 0

pH = 3.46 + log(0) = undefined

Therefore, before the addition of any KOH, the pH is undefined.

Note: Since HF is a weak acid and the initial concentration is not too small (0.20 M), the assumption that [A-] = 0 is valid.

Question 13

A 100.0 mL sample of 0.20 M HF is titrated with 0.10 M KOH.Determine the pH of the solution after the addition of 100.0 mL of KOH.The K_a of HF is 3.5 × 10^-4.

Accepted Answer

First, we need to determine the amount of moles of HF in the initial solution:

mol HF = concentration x volume in liters
mol HF = 0.20 M x 0.100 L
mol HF = 0.020

Next, we need to determine how much KOH is required to completely neutralize the HF.

1 mol KOH reacts with 1 mol HF, so the amount of moles of KOH required is also 0.020.

Now, we need to determine how much KOH is added during the titration. We know that 100.0 mL of 0.10 M KOH is added, so we can determine the amount of moles of KOH added:

mol KOH = concentration x volume in liters
mol KOH = 0.10 M x 0.100 L
mol KOH = 0.010

Since this is less than the amount of KOH required to fully neutralize the HF, we know that there will be some KOH left over after the titration.

The reaction between HF and KOH produces KF and water:

HF + KOH -> KF + H2O

Since KOH is a strong base, it will react completely with water to produce OH- ions:

KOH + H2O -> K+ + OH- + H2O

The OH- ions from the excess KOH will react with the HF to produce F- ions and water:

HF + OH- -> F- + H2O

Since the initial solution contains 0.020 moles of HF and the amount of KOH added only neutralizes half of this, we know that there will be 0.010 moles of HF left over at the end of the reaction. This means that there will be 0.010 moles of F- ions produced.

We can use an ICE table to determine the concentration of the F- ions:

I: HF + OH- -> F- + H2O
C: -x + x -> x
E: 0.010 - x x x

We know that the equilibrium constant (Ka) for HF is 3.5 x 10^-4, so we can set up an expression for the equilibrium constant:

Ka = [F-][H3O+]/[HF]
3.5 x 10^-4 = x^2/(0.020 - x)

Since the initial concentration of HF is much larger than the concentration of F- at equilibrium, we can assume that the change in the concentration of HF is approximately equal to the initial concentration of HF. This allows us to simplify the expression:

3.5 x 10^-4 = x^2/0.020

Solving for x gives us:

x = 4.70 x 10^-3

To find the pH, we need to determine the concentration of H3O+:

[H3O+] = Ka[H2F+]/[HF]
[H3O+] = (3.5 x 10^-4)(4.70 x 10^-3)/(0.020 - 4.70 x 10^-3)
[H3O+] = 9.15 x 10^-4

pH = -log[H3O+]
pH = -log(9.15 x 10^-4)
pH = 3.46

Therefore, the answer is D) 3.46.

Question 14

Which of the following compounds will be more soluble in acidic solution than in pure water?

Accepted Answer

FeS is more soluble in acidic solution because the sulfide ion (S^2-) reacts with H^+ ions to form H2S gas, which reduces the concentration of sulfide ions and shifts the equilibrium towards dissolution.

Question 15

A 100.0 mL sample of 0.20 M HF is titrated with 0.10 M KOH.Determine the pH of the solution after the addition of 200.0 mL of KOH.The K_a of HF is 3.5 × 10^-4.

Accepted Answer

The answer of A 100.0 mL sample of 0.20 M...

Question 16

A 100.0 mL sample of 0.10 M Ca(OH)₂ is titrated with 0.10 M HBr.Determine the pH of the solution after the addition of 100.0 mL HBr.

Accepted Answer

The answer of A 100.0 mL sample of 0.10 M...

Question 17

Give the expression for the solubility product constant for PbCl₂.

Accepted Answer

The answer of Give the expression for the solubility product...

Question 18

A 100.0 mL sample of 0.10 M NH₃ is titrated with 0.10 M HNO₃.Determine the pH of the solution after the addition of 200.0 mL of HNO₃.The K_b of NH₃ is 1.8 × 10^-5.

Accepted Answer

The answer of A 100.0 mL sample of 0.10 M...

Question 19

A 100.0 mL sample of 0.10 M Ca(OH)₂ is titrated with 0.10 M HBr.Determine the pH of the solution after the addition of 300.0 mL HBr.

Accepted Answer

After adding 300.0 mL of 0.10 M HBr to 100.0 mL of 0.10 M Ca(OH)2, there is an excess of HBr. The moles of HBr added are 0.030 mol, and the moles of Ca(OH)2 initially present are 0.010 mol. The reaction neutralizes 0.020 mol of HBr, leaving 0.010 mol of HBr in 400.0 mL of solution. The concentration of HBr is 0.025 M, leading to a pH of approximately 1.60.

Question 20

A 100.0 mL sample of 0.10 M Ca(OH)₂ is titrated with 0.10 M HBr.Determine the pH of the solution after the addition of 200.0 mL HBr.

Accepted Answer

The answer of A 100.0 mL sample of 0.10 M...

Quiz 16: Aqueous Ionic Equilibrium

A 100.0 mL sample of 0.10 M Ca(OH) ₂ is titrated with 0.10 M HBr.Determine the pH of the solution after the addition of 400.0 mL HBr.

Which of the following compounds solubility will not be affected by a low pH in solution?

A 100.0 mL sample of 0.10 M NH₃ is titrated with 0.10 M HNO₃.Determine the pH of the solution after the addition of 50.0 mL of HNO₃.The K_b of NH₃ is 1.8 × 10^-5.

A 100.0 mL sample of 0.10 M NH₃ is titrated with 0.10 M HNO₃.Determine the pH of the solution after the addition of 100.0 mL of HNO₃.The K_b of NH₃ is 1.8 × 10^-5.

A 100.0 mL sample of 0.18 M HClO₄ is titrated with 0.27 M LiOH.Determine the pH of the solution after the addition of 100.0 mL of LiOH.

A 100.0 mL sample of 0.10 M Ca(OH) ₂ is titrated with 0.10 M HBr.Determine the pH of the solution before the addition of any HBr.

A 100.0 mL sample of 0.20 M HF is titrated with 0.10 M KOH.Determine the pH of the solution after the addition of 300.0 mL of KOH.The K_a of HF is 3.5 × 10^-4.

Which of the following compounds will have the highest molar solubility in pure water?

A 100.0 mL sample of 0.10 M NH₃ is titrated with 0.10 M HNO₃.Determine the pH of the solution before the addition of any HNO₃.The K_b of NH₃ is 1.8 × 10^-5.

A 100.0 mL sample of 0.10 M NH₃ is titrated with 0.10 M HNO₃.Determine the pH of the solution after the addition of 150.0 mL of HNO₃.The K_b of NH₃ is 1.8 × 10^-5.

A 100.0 mL sample of 0.20 M HF is titrated with 0.10 M KOH.Determine the pH of the solution after the addition of 400.0 mL of KOH.The K_a of HF is 3.5 × 10^-4.

A 100.0 mL sample of 0.20 M HF is titrated with 0.10 M KOH.Determine the pH of the solution before the addition of any KOH.The K_a of HF is 3.5 × 10^-4.

A 100.0 mL sample of 0.20 M HF is titrated with 0.10 M KOH.Determine the pH of the solution after the addition of 100.0 mL of KOH.The K_a of HF is 3.5 × 10^-4.

Which of the following compounds will be more soluble in acidic solution than in pure water?

A 100.0 mL sample of 0.20 M HF is titrated with 0.10 M KOH.Determine the pH of the solution after the addition of 200.0 mL of KOH.The K_a of HF is 3.5 × 10^-4.

A 100.0 mL sample of 0.10 M Ca(OH) ₂ is titrated with 0.10 M HBr.Determine the pH of the solution after the addition of 100.0 mL HBr.

Give the expression for the solubility product constant for PbCl₂.

A 100.0 mL sample of 0.10 M NH₃ is titrated with 0.10 M HNO₃.Determine the pH of the solution after the addition of 200.0 mL of HNO₃.The K_b of NH₃ is 1.8 × 10^-5.

A 100.0 mL sample of 0.10 M Ca(OH) ₂ is titrated with 0.10 M HBr.Determine the pH of the solution after the addition of 300.0 mL HBr.

A 100.0 mL sample of 0.10 M Ca(OH) ₂ is titrated with 0.10 M HBr.Determine the pH of the solution after the addition of 200.0 mL HBr.

Quiz 16: Aqueous Ionic Equilibrium

A 100.0 mL sample of 0.10 M Ca(OH) 2 is titrated with 0.10 M HBr.Determine the pH of the solution after the addition of 400.0 mL HBr.

Which of the following compounds solubility will not be affected by a low pH in solution?

A 100.0 mL sample of 0.10 M NH3 is titrated with 0.10 M HNO3.Determine the pH of the solution after the addition of 50.0 mL of HNO3.The Kb of NH3 is 1.8 × 10-5.

A 100.0 mL sample of 0.10 M NH3 is titrated with 0.10 M HNO3.Determine the pH of the solution after the addition of 100.0 mL of HNO3.The Kb of NH3 is 1.8 × 10-5.

A 100.0 mL sample of 0.18 M HClO4 is titrated with 0.27 M LiOH.Determine the pH of the solution after the addition of 100.0 mL of LiOH.

A 100.0 mL sample of 0.10 M Ca(OH) 2 is titrated with 0.10 M HBr.Determine the pH of the solution before the addition of any HBr.

A 100.0 mL sample of 0.20 M HF is titrated with 0.10 M KOH.Determine the pH of the solution after the addition of 300.0 mL of KOH.The Ka of HF is 3.5 × 10-4.

Which of the following compounds will have the highest molar solubility in pure water?

A 100.0 mL sample of 0.10 M NH3 is titrated with 0.10 M HNO3.Determine the pH of the solution before the addition of any HNO3.The Kb of NH3 is 1.8 × 10-5.

A 100.0 mL sample of 0.10 M NH3 is titrated with 0.10 M HNO3.Determine the pH of the solution after the addition of 150.0 mL of HNO3.The Kb of NH3 is 1.8 × 10-5.

A 100.0 mL sample of 0.20 M HF is titrated with 0.10 M KOH.Determine the pH of the solution after the addition of 400.0 mL of KOH.The Ka of HF is 3.5 × 10-4.

A 100.0 mL sample of 0.20 M HF is titrated with 0.10 M KOH.Determine the pH of the solution before the addition of any KOH.The Ka of HF is 3.5 × 10-4.

A 100.0 mL sample of 0.20 M HF is titrated with 0.10 M KOH.Determine the pH of the solution after the addition of 100.0 mL of KOH.The Ka of HF is 3.5 × 10-4.

Which of the following compounds will be more soluble in acidic solution than in pure water?

A 100.0 mL sample of 0.20 M HF is titrated with 0.10 M KOH.Determine the pH of the solution after the addition of 200.0 mL of KOH.The Ka of HF is 3.5 × 10-4.

A 100.0 mL sample of 0.10 M Ca(OH) 2 is titrated with 0.10 M HBr.Determine the pH of the solution after the addition of 100.0 mL HBr.

Give the expression for the solubility product constant for PbCl2.

A 100.0 mL sample of 0.10 M NH3 is titrated with 0.10 M HNO3.Determine the pH of the solution after the addition of 200.0 mL of HNO3.The Kb of NH3 is 1.8 × 10-5.

A 100.0 mL sample of 0.10 M Ca(OH) 2 is titrated with 0.10 M HBr.Determine the pH of the solution after the addition of 300.0 mL HBr.

A 100.0 mL sample of 0.10 M Ca(OH) 2 is titrated with 0.10 M HBr.Determine the pH of the solution after the addition of 200.0 mL HBr.

A 100.0 mL sample of 0.10 M Ca(OH) ₂ is titrated with 0.10 M HBr.Determine the pH of the solution after the addition of 400.0 mL HBr.

A 100.0 mL sample of 0.10 M NH₃ is titrated with 0.10 M HNO₃.Determine the pH of the solution after the addition of 50.0 mL of HNO₃.The K_b of NH₃ is 1.8 × 10^-5.

A 100.0 mL sample of 0.10 M NH₃ is titrated with 0.10 M HNO₃.Determine the pH of the solution after the addition of 100.0 mL of HNO₃.The K_b of NH₃ is 1.8 × 10^-5.

A 100.0 mL sample of 0.18 M HClO₄ is titrated with 0.27 M LiOH.Determine the pH of the solution after the addition of 100.0 mL of LiOH.

A 100.0 mL sample of 0.10 M Ca(OH) ₂ is titrated with 0.10 M HBr.Determine the pH of the solution before the addition of any HBr.

A 100.0 mL sample of 0.20 M HF is titrated with 0.10 M KOH.Determine the pH of the solution after the addition of 300.0 mL of KOH.The K_a of HF is 3.5 × 10^-4.

A 100.0 mL sample of 0.10 M NH₃ is titrated with 0.10 M HNO₃.Determine the pH of the solution before the addition of any HNO₃.The K_b of NH₃ is 1.8 × 10^-5.

A 100.0 mL sample of 0.10 M NH₃ is titrated with 0.10 M HNO₃.Determine the pH of the solution after the addition of 150.0 mL of HNO₃.The K_b of NH₃ is 1.8 × 10^-5.

A 100.0 mL sample of 0.20 M HF is titrated with 0.10 M KOH.Determine the pH of the solution after the addition of 400.0 mL of KOH.The K_a of HF is 3.5 × 10^-4.

A 100.0 mL sample of 0.20 M HF is titrated with 0.10 M KOH.Determine the pH of the solution before the addition of any KOH.The K_a of HF is 3.5 × 10^-4.

A 100.0 mL sample of 0.20 M HF is titrated with 0.10 M KOH.Determine the pH of the solution after the addition of 100.0 mL of KOH.The K_a of HF is 3.5 × 10^-4.

A 100.0 mL sample of 0.20 M HF is titrated with 0.10 M KOH.Determine the pH of the solution after the addition of 200.0 mL of KOH.The K_a of HF is 3.5 × 10^-4.

A 100.0 mL sample of 0.10 M Ca(OH) ₂ is titrated with 0.10 M HBr.Determine the pH of the solution after the addition of 100.0 mL HBr.

Give the expression for the solubility product constant for PbCl₂.

A 100.0 mL sample of 0.10 M NH₃ is titrated with 0.10 M HNO₃.Determine the pH of the solution after the addition of 200.0 mL of HNO₃.The K_b of NH₃ is 1.8 × 10^-5.

A 100.0 mL sample of 0.10 M Ca(OH) ₂ is titrated with 0.10 M HBr.Determine the pH of the solution after the addition of 300.0 mL HBr.

A 100.0 mL sample of 0.10 M Ca(OH) ₂ is titrated with 0.10 M HBr.Determine the pH of the solution after the addition of 200.0 mL HBr.