Question 1

A solution is formed by mixing 50.0 mL of 10.00 M NaCN with 50.0 mL of 2.0 × 10S1U1P1-3S1S1P0 M CuNOS1U1B13S1U1B0. Cu(I) forms complex ions with cyanide as follows: Calculate the following concentrations at equilibrium: -[Cu⁺]

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Question 2

Calculate the pH of the final solution obtained by mixing the following solutions. For HCN, K_a = 6.2 × 10^-10.50.0 mL of 0.10 M HNO₃ 60.0 mL of 0.20 M Ba(OH)₂ 95.0 mL of 0.20 M HClO₄195.0 mL of 1.0 × 10^-4 M HCN

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Question 3

Explain how to decide on a specific indicator for a given titration.

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Question 4

A 50.0-mL sample of 2.0 × 10^-4 M CuNO₃ is added to 50.0 mL of 4.0 M NaCN. Cu⁺ reacts with CN^- to form the complex ion Cu(CN)₃^2-: The concentration of Cu⁺ at equilibrium is

Accepted Answer

First, we can write the balanced chemical equation for the reaction: CuNO₃⁺ + NaCN^- → Cu(CN)₃^2- + NaNO₃⁺ The reaction is an equilibrium reaction, and we are given the initial concentrations of CuNO₃⁺ and NaCN, both at 50.0 mL. We can use an ICE table (initial, change, equilibrium) to solve for the concentration of Cu⁺ at equilibrium. | |CuNO₃⁺ | NaCN | Cu(CN)₃^2- | |--------|-------------------|--------|------------------------| |Initial | 2.0×10^-4 M | 4.0 M | 0 M | |Change | -x | -x/2 | +x | |Equilibrium|2.0×10^-4 M - x | 4.0 M - x/2 | x | At equilibrium, the concentration of Cu(CN)₃^2- is equal to the concentration of CuNO₃⁺ that reacted to form the complex, which is x. To solve for x, we need to use the equilibrium constant expression for the reaction. The equilibrium constant (K) is given as: K = [Cu(CN)₃^2-] / ([Cu⁺][NaCN^-]/2) Substituting in the equilibrium concentrations: K = x / [(2.0×10^-4 - x)(4.0 - x/2)] We are given that K = 6.0×10^-28. Rearranging and solving for x, we get: x = (K[NaCN]/2) / (1 + K[NaCN]/2 + K[NaCN]([CuNO₃⁺]/[Cu⁺])) Substituting in values: x = (6.0×10^-28 × 2.0) / (1 + 6.0×10^-28 × 2.0 + 6.0×10^-28 × (2.0×10^-4 / [Cu⁺])) x ≈ 1.2×10^-14 M Therefore, the concentration of Cu⁺ at equilibrium is approximately 1.2×10^-14 M, which is Choice (B).

Question 5

The value of K_f for the complex ion Ag(NH₃)₂⁺ is 1.7 × 10⁷. K_sp for AgCl is 1.6 × 10^-10. Calculate the molar solubility of AgCl in 1.0 M NH₃.

Accepted Answer

Step 1: Write the solubility product expression:

AgCl(s) ⇌ Ag+(aq) + Cl-(aq)

Ksp = [Ag+][Cl-]

Step 2: Write the formation constant expression for the complex ion:

Ag(NHS13-)(aq) + 12S2-(aq) ⇌ Ag(NHS13-)(S2)12(aq)

Kf = [Ag(NHS13-)(S2)12] / [Ag(NHS13-)][S2-]12

Step 3: Write the overall equilibrium expression for the dissolution of AgCl in the presence of Ag(NHS13-) and S2-:

AgCl(s) + Ag(NHS13-)(aq) + 12S2-(aq) ⇌ Ag(NHS13-)(S2)12(aq) + Cl-(aq) + Ag+(aq)

Koverall = [Ag(NHS13-)(S2)12][Ag+][Cl-] / [Ag(NHS13-)][S2-]12[AgCl(s)]

Step 4: Use the values given in the problem to solve for the molar solubility of AgCl:

Ksp = 1.6 × 10-10
Kf = 1.7 × 1017
[NHS13-] = 1.0 M
[S2-] = 1.0 M

Koverall = (1.7 × 1017)(1.0 M)(1.0 M)12 / (1.0 M)(1.0 M)12(1.6 × 10-10) = 4.7 × 10-12

Solving for [Ag+] and [Cl-], assuming they are equal because AgCl is a 1:1 electrolyte:

Ksp = [Ag+]2
[Ag+] = [Cl-] = (1.6 × 10-10)1/2 = 1.3 × 10-5

Substituting into the overall equilibrium expression:

4.7 × 10-12 = (1.7 × 1017)(1.3 × 10-5)2 / (1.0 M)(1.0 M)12[S]

[S] = 4.7 × 10-2 M = 4.7 × 10-3 mol/L

Therefore, the molar solubility of AgCl in 1.0 M NHS13- is 4.7 × 10-3 mol/L.

The best choice is C, as it is the only option with a value close to 4.7 × 10-3 mol/L.

Question 6

Consider a solution made by mixing 500.0 mL of 4.0 M NH₃ and 500.0 mL of 0.40 M AgNO₃. Ag⁺ reacts with NH₃ to form AgNH₃⁺ and Ag(NH₃)₂⁺: The concentration of Ag⁺ at equilibrium is

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Question 7

Calculate the molar concentration of uncomplexed Zn²⁺ in a solution that contains 0.20 mol of Zn(NH₃)₄²⁺ per liter. The overall K_f for Zn(NH₃)₄²⁺ is 3.8 × 10⁹.

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To solve this problem, we need to use the equilibrium constant expression: K_f = [Zn²⁺] / [Zn(NH₃)₄²⁺] We know the value of K_f and we are given the concentration of Zn(NH₃)₄²⁺, so our task is to find [Zn²⁺]. To do this, we can use the fact that the total concentration of zinc in the solution will be: [zinc]total = [Zn²⁺] + [Zn(NH₃)₄²⁺] We know that [Zn(NH₃)₄²⁺] = 0.20 mol/L, so we just need to solve for [Zn²⁺] in the equation above. Rearranging, we get: [Zn²⁺] = [zinc]total - [Zn(NH₃)₄²⁺] [Zn²⁺] = (0.20 mol/L) - [Zn(NH₃)₄²⁺] Now we can substitute this into our equilibrium constant expression and solve for [zinc]total: 3.8 × 10⁹ = [(0.20 mol/L) - [Zn(NH₃)₄²⁺]] / [Zn(NH₃)₄²⁺] Simplifying: [Zn(NH₃)₄²⁺] = 0.20 mol/L [Zn²⁺] = 0.152 mol/L [zinc]total = 0.352 mol/L Finally, we can calculate the molar concentration of uncomplexed Zn²⁺: [Zn²⁺] = 0.152 mol/L Total volume of solution = 1 L Molarity of Zn²⁺ = [Zn²⁺] / total volume = 0.152 mol/L Therefore, the answer is B) 2.9 × 10^-3 M.

Question 8

Derive the equation describing the relationship between the pH at the first equivalence point and the K_a values of the diprotic weak acid being titrated with NaOH.

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Question 9

A 50.0-mL sample of 2.0 × 10^-4 M CuNO₃ is added to 50.0 mL of 4.0 M NaCN. Cu⁺ reacts with CN^- to form the complex ion Cu(CN)₃^2-: What is the concentration of CN^- at equilibrium?

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Question 10

Explain why the pH of an aqueous solution of NaHCO₃ is independent of concentration.

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Question 11

Given the following values of equilibrium constants: What is the value of the equilibrium constant for the following reaction? Cu(OH)₂(s) + 4NH₃(aq) Cu(NH₃)₄²⁺(aq) + 2OH^-(aq)

Accepted Answer

The equilibrium constant expression for the given reaction is:

K = (Cu(NH$_3$)$_4$(OH)$_2^{2+}$)(OH$^-$)$_2$ / (Cu(OH)$_2$)(NH$_3$)$_4$(OH)$_2^{2+}$)(NH$_3$)$_2$

The given equilibrium constants can be used to form a series of equations that relate K for the given reaction to the given equilibrium constants:

K$_1$ = (Cu(OH)$_2$)/(Cu$^{2+}$)(OH$^-$)$^2$
K$_2$ = (Cu(NH$_3$)$_4$(OH)$_2^{2+}$)/(Cu$^{2+}$)(OH$^-$)$^2$(NH$_3$)$_2$
K$_3$ = (Cu(NH$_3$)$_4^{2+}$)/(Cu$^{2+}$)(NH$_3$)$_4$
K$_4$ = (OH$^-$)$_2$/ (H$_2$O)

By multiplying these equations together and simplifying, we get:

K = K$_2$ / K$_1$ = K$_3$(OH$^-$)$_2$/ (Cu(OH)$_2$)(NH$_3$)$_2$

Plugging in the given values for the equilibrium constants, we get:

K = (1.6 × 10$^{-6}$)(10$^{-4}$)$^2$ / (1.1 × 10$^{-18}$)(10$^{-4}$)$^2$ = 1.6 × 10$^{-6}$

Therefore, the correct answer is choice B, 1.6 × 10$^{-6}$.

Question 12

Differentiate between the equivalence point and the endpoint in an acid-base titration.

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Question 13

Derive the Henderson-Hasselbalch equation from the K_a expression. Explain the assumptions inherent in the Henderson-Hasselbalch equation.

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Question 14

A solution is formed by mixing 50.0 mL of 10.00 M NaCN with 50.0 mL of 2.0 × 10S1U1P1-3S1S1P0 M CuNOS1U1B13S1U1B0. Cu(I) forms complex ions with cyanide as follows: Calculate the following concentrations at equilibrium: -[Cu(CN)₃^2-]

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Question 15

A 50.0-mL solution of the acid H₃A (K_a1 = 1.0 × 10^-6 , K_a2 = 1.0 × 10^-9, and K_a3 = 1.0 × 10^-12.) is titrated with 0.050 M KOH. The second equivalence point is reached at 25.0 mL of base. Calculate the original concentration of H₃A and the volume of 0.050 M KOH needed to reach a pH of 6.70.

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Question 16

A 50.0-mL sample of 2.0 × 10^-4 M CuNO₃ is added to 50.0 mL of 4.0 M NaCN. Cu⁺ reacts with CN^- to form the complex ion Cu(CN)₃^2-: Calculate the solubility of CuBr(s) (K_sp = 1.0 × 10^-5) in 1.0 L of 1.0 M NaCN.

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Question 17

A solution is formed by mixing 50.0 mL of 10.00 M NaCN with 50.0 mL of 2.0 × 10S1U1P1-3S1S1P0 M CuNOS1U1B13S1U1B0. Cu(I) forms complex ions with cyanide as follows: Calculate the following concentrations at equilibrium: -[Cu(CN)₂^-]

Accepted Answer

First, let's write out the balanced equation for the reaction:

Cu+ + CN- --> Cu(CN)2-

We start with 50.0 mL of 10.00 M NaCN, which contains 0.50 moles of CN-. We also start with 50.0 mL of 2.0 × 10^-3 M Cu+, which contains 1.0 × 10^-4 moles of Cu+.

The reaction between Cu+ and CN- will consume some of the CN-, forming the Cu(CN)2- complex ion. Let's define x as the amount of CN- that reacts to form the complex. Then:

Cu+ + x CN- --> Cu(CN)2-

We can set up an ICE (initial, change, equilibrium) table:

Cu+         CN-          Cu(CN)2- 
I        1.0 × 10^-4   0.50          0 
C       -x            -x            +x 
E        1.0 × 10^-4 - x  0.50 - x   x

We can use the Kf expression to solve for x:

Kf = [Cu(CN)2-]/([Cu+][CN-]^2)

Kf = 4.8 × 10^20 M^-2 from the given data

Kf = x/([1.0 × 10^-4][0.50 - x]^2)

Simplifying:

4.8 × 10^20 = x/(1.25 × 10^-5 - 1.5x + x^2)

We can assume that x is small compared to 0.50, so we can simplify further:

4.8 × 10^20 = x/(1.25 × 10^-5)

x = 6.0 × 10^15 M

Now we can calculate the concentration of Cu(CN)2- at equilibrium:

[Cu(CN)2-] = x/0.05 L = 1.2 × 10^-13 M

(Note that the volume of the final solution is 100.0 mL.)

So the answer is (C) 2.0 × 10^-7 M.

Question 18

You have 0.20 M HNO₂ (K_a = 4.0 × 10^-4) and 0.20 M KNO₂. You need 1.00 L of a buffered solution at a pH of 3.00. What volumes of each solution do you add together to make this buffered solution?

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Question 19

The cation M²⁺ reacts with NH₃ to form a series of complex ions as follows: A 1.0 × 10^-3 mol sample of M(NO₃)₂ is added to 1.0 L of 15.0 M NH₃ (K_b = 1.8 × 10^-5). Which of the following would be a dominant species in this solution?

Accepted Answer

Since the concentration of the NH₃ is much higher than the concentration of the M(NO₃)₂ (10^3 times), the reaction will favor the formation of the product, which has a higher concentration of NH₃. Therefore, the dominant species in this solution would be the product with maximum concentration of NH₃, which is M(NH₃)₃²⁺.

Question 20

Explain why we cannot directly compare K_sp values in comparing solubilities of ionic solids with different numbers of ions.

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Quiz 8: Applications of Aqueous Equilibria

A solution is formed by mixing 50.0 mL of 10.00 M NaCN with 50.0 mL of 2.0 × 10^-3 M CuNO₃. Cu(I) forms complex ions with cyanide as follows: Calculate the following concentrations at equilibrium: -[Cu⁺]

Calculate the pH of the final solution obtained by mixing the following solutions. For HCN, K_a = 6.2 × 10^-10.50.0 mL of 0.10 M HNO₃ 60.0 mL of 0.20 M Ba(OH)₂ 95.0 mL of 0.20 M HClO₄195.0 mL of 1.0 × 10^-4 M HCN

Explain how to decide on a specific indicator for a given titration.

A 50.0-mL sample of 2.0 × 10^-4 M CuNO₃ is added to 50.0 mL of 4.0 M NaCN. Cu⁺ reacts with CN^- to form the complex ion Cu(CN) ₃^2-: The concentration of Cu⁺ at equilibrium is

The value of K_f for the complex ion Ag(NH₃) ₂⁺ is 1.7 × 10⁷. K_sp for AgCl is 1.6 × 10^-10. Calculate the molar solubility of AgCl in 1.0 M NH₃.

Consider a solution made by mixing 500.0 mL of 4.0 M NH₃ and 500.0 mL of 0.40 M AgNO₃. Ag⁺ reacts with NH₃ to form AgNH₃⁺ and Ag(NH₃) ₂⁺: The concentration of Ag⁺ at equilibrium is

Calculate the molar concentration of uncomplexed Zn²⁺ in a solution that contains 0.20 mol of Zn(NH₃) ₄²⁺ per liter. The overall K_f for Zn(NH₃) ₄²⁺ is 3.8 × 10⁹.

Derive the equation describing the relationship between the pH at the first equivalence point and the K_a values of the diprotic weak acid being titrated with NaOH.

A 50.0-mL sample of 2.0 × 10^-4 M CuNO₃ is added to 50.0 mL of 4.0 M NaCN. Cu⁺ reacts with CN^- to form the complex ion Cu(CN) ₃^2-: What is the concentration of CN^- at equilibrium?

Explain why the pH of an aqueous solution of NaHCO₃ is independent of concentration.

Given the following values of equilibrium constants: What is the value of the equilibrium constant for the following reaction? Cu(OH) ₂(s) + 4NH₃(aq) Cu(NH₃) ₄²⁺(aq) + 2OH^-(aq)

Differentiate between the equivalence point and the endpoint in an acid-base titration.

Derive the Henderson-Hasselbalch equation from the K_a expression. Explain the assumptions inherent in the Henderson-Hasselbalch equation.

A solution is formed by mixing 50.0 mL of 10.00 M NaCN with 50.0 mL of 2.0 × 10^-3 M CuNO₃. Cu(I) forms complex ions with cyanide as follows: Calculate the following concentrations at equilibrium: -[Cu(CN) ₃^2-]

A 50.0-mL sample of 2.0 × 10^-4 M CuNO₃ is added to 50.0 mL of 4.0 M NaCN. Cu⁺ reacts with CN^- to form the complex ion Cu(CN) ₃^2-: Calculate the solubility of CuBr(s) (K_sp = 1.0 × 10^-5) in 1.0 L of 1.0 M NaCN.

A solution is formed by mixing 50.0 mL of 10.00 M NaCN with 50.0 mL of 2.0 × 10^-3 M CuNO₃. Cu(I) forms complex ions with cyanide as follows: Calculate the following concentrations at equilibrium: -[Cu(CN) ₂^-]

You have 0.20 M HNO₂ (K_a = 4.0 × 10^-4) and 0.20 M KNO₂. You need 1.00 L of a buffered solution at a pH of 3.00. What volumes of each solution do you add together to make this buffered solution?

The cation M²⁺ reacts with NH₃ to form a series of complex ions as follows: A 1.0 × 10^-3 mol sample of M(NO₃) ₂ is added to 1.0 L of 15.0 M NH₃ (K_b = 1.8 × 10^-5) . Which of the following would be a dominant species in this solution?

Explain why we cannot directly compare K_sp values in comparing solubilities of ionic solids with different numbers of ions.

Quiz 8: Applications of Aqueous Equilibria

A solution is formed by mixing 50.0 mL of 10.00 M NaCN with 50.0 mL of 2.0 × 10-3 M CuNO3. Cu(I) forms complex ions with cyanide as follows: Calculate the following concentrations at equilibrium: -[Cu+]

Calculate the pH of the final solution obtained by mixing the following solutions. For HCN, Ka = 6.2 × 10-10.50.0 mL of 0.10 M HNO3 60.0 mL of 0.20 M Ba(OH)2 95.0 mL of 0.20 M HClO4 195.0 mL of 1.0 × 10-4 M HCN

Explain how to decide on a specific indicator for a given titration.

A 50.0-mL sample of 2.0 × 10-4 M CuNO3 is added to 50.0 mL of 4.0 M NaCN. Cu+ reacts with CN- to form the complex ion Cu(CN) 32-: The concentration of Cu+ at equilibrium is

The value of Kf for the complex ion Ag(NH3) 2+ is 1.7 × 107. Ksp for AgCl is 1.6 × 10-10. Calculate the molar solubility of AgCl in 1.0 M NH3.

Consider a solution made by mixing 500.0 mL of 4.0 M NH3 and 500.0 mL of 0.40 M AgNO3. Ag+ reacts with NH3 to form AgNH3+ and Ag(NH3) 2+: The concentration of Ag+ at equilibrium is

Calculate the molar concentration of uncomplexed Zn2+ in a solution that contains 0.20 mol of Zn(NH3) 42+ per liter. The overall Kf for Zn(NH3) 42+ is 3.8 × 109.

Derive the equation describing the relationship between the pH at the first equivalence point and the Ka values of the diprotic weak acid being titrated with NaOH.

A 50.0-mL sample of 2.0 × 10-4 M CuNO3 is added to 50.0 mL of 4.0 M NaCN. Cu+ reacts with CN- to form the complex ion Cu(CN) 32-: What is the concentration of CN- at equilibrium?

Explain why the pH of an aqueous solution of NaHCO3 is independent of concentration.

Given the following values of equilibrium constants: What is the value of the equilibrium constant for the following reaction? Cu(OH) 2(s) + 4NH3(aq) Cu(NH3) 42+(aq) + 2OH-(aq)

Differentiate between the equivalence point and the endpoint in an acid-base titration.

Derive the Henderson-Hasselbalch equation from the Ka expression. Explain the assumptions inherent in the Henderson-Hasselbalch equation.

A solution is formed by mixing 50.0 mL of 10.00 M NaCN with 50.0 mL of 2.0 × 10-3 M CuNO3. Cu(I) forms complex ions with cyanide as follows: Calculate the following concentrations at equilibrium: -[Cu(CN) 32-]

A 50.0-mL solution of the acid H3A (Ka1 = 1.0 × 10-6 , Ka2 = 1.0 × 10-9, and Ka3 = 1.0 × 10-12.) is titrated with 0.050 M KOH. The second equivalence point is reached at 25.0 mL of base. Calculate the original concentration of H3A and the volume of 0.050 M KOH needed to reach a pH of 6.70.

A 50.0-mL sample of 2.0 × 10-4 M CuNO3 is added to 50.0 mL of 4.0 M NaCN. Cu+ reacts with CN- to form the complex ion Cu(CN) 32-: Calculate the solubility of CuBr(s) (Ksp = 1.0 × 10-5) in 1.0 L of 1.0 M NaCN.

A solution is formed by mixing 50.0 mL of 10.00 M NaCN with 50.0 mL of 2.0 × 10-3 M CuNO3. Cu(I) forms complex ions with cyanide as follows: Calculate the following concentrations at equilibrium: -[Cu(CN) 2-]

You have 0.20 M HNO2 (Ka = 4.0 × 10-4) and 0.20 M KNO2. You need 1.00 L of a buffered solution at a pH of 3.00. What volumes of each solution do you add together to make this buffered solution?

The cation M2+ reacts with NH3 to form a series of complex ions as follows: A 1.0 × 10-3 mol sample of M(NO3) 2 is added to 1.0 L of 15.0 M NH3 (Kb = 1.8 × 10-5) . Which of the following would be a dominant species in this solution?

Explain why we cannot directly compare Ksp values in comparing solubilities of ionic solids with different numbers of ions.

A solution is formed by mixing 50.0 mL of 10.00 M NaCN with 50.0 mL of 2.0 × 10^-3 M CuNO₃. Cu(I) forms complex ions with cyanide as follows: Calculate the following concentrations at equilibrium: -[Cu⁺]

Calculate the pH of the final solution obtained by mixing the following solutions. For HCN, K_a = 6.2 × 10^-10.50.0 mL of 0.10 M HNO₃ 60.0 mL of 0.20 M Ba(OH)₂ 95.0 mL of 0.20 M HClO₄195.0 mL of 1.0 × 10^-4 M HCN

A 50.0-mL sample of 2.0 × 10^-4 M CuNO₃ is added to 50.0 mL of 4.0 M NaCN. Cu⁺ reacts with CN^- to form the complex ion Cu(CN) ₃^2-: The concentration of Cu⁺ at equilibrium is

The value of K_f for the complex ion Ag(NH₃) ₂⁺ is 1.7 × 10⁷. K_sp for AgCl is 1.6 × 10^-10. Calculate the molar solubility of AgCl in 1.0 M NH₃.

Consider a solution made by mixing 500.0 mL of 4.0 M NH₃ and 500.0 mL of 0.40 M AgNO₃. Ag⁺ reacts with NH₃ to form AgNH₃⁺ and Ag(NH₃) ₂⁺: The concentration of Ag⁺ at equilibrium is

Calculate the molar concentration of uncomplexed Zn²⁺ in a solution that contains 0.20 mol of Zn(NH₃) ₄²⁺ per liter. The overall K_f for Zn(NH₃) ₄²⁺ is 3.8 × 10⁹.

Derive the equation describing the relationship between the pH at the first equivalence point and the K_a values of the diprotic weak acid being titrated with NaOH.

A 50.0-mL sample of 2.0 × 10^-4 M CuNO₃ is added to 50.0 mL of 4.0 M NaCN. Cu⁺ reacts with CN^- to form the complex ion Cu(CN) ₃^2-: What is the concentration of CN^- at equilibrium?

Explain why the pH of an aqueous solution of NaHCO₃ is independent of concentration.

Given the following values of equilibrium constants: What is the value of the equilibrium constant for the following reaction? Cu(OH) ₂(s) + 4NH₃(aq) Cu(NH₃) ₄²⁺(aq) + 2OH^-(aq)

Derive the Henderson-Hasselbalch equation from the K_a expression. Explain the assumptions inherent in the Henderson-Hasselbalch equation.

A solution is formed by mixing 50.0 mL of 10.00 M NaCN with 50.0 mL of 2.0 × 10^-3 M CuNO₃. Cu(I) forms complex ions with cyanide as follows: Calculate the following concentrations at equilibrium: -[Cu(CN) ₃^2-]

A 50.0-mL sample of 2.0 × 10^-4 M CuNO₃ is added to 50.0 mL of 4.0 M NaCN. Cu⁺ reacts with CN^- to form the complex ion Cu(CN) ₃^2-: Calculate the solubility of CuBr(s) (K_sp = 1.0 × 10^-5) in 1.0 L of 1.0 M NaCN.

A solution is formed by mixing 50.0 mL of 10.00 M NaCN with 50.0 mL of 2.0 × 10^-3 M CuNO₃. Cu(I) forms complex ions with cyanide as follows: Calculate the following concentrations at equilibrium: -[Cu(CN) ₂^-]

You have 0.20 M HNO₂ (K_a = 4.0 × 10^-4) and 0.20 M KNO₂. You need 1.00 L of a buffered solution at a pH of 3.00. What volumes of each solution do you add together to make this buffered solution?

The cation M²⁺ reacts with NH₃ to form a series of complex ions as follows: A 1.0 × 10^-3 mol sample of M(NO₃) ₂ is added to 1.0 L of 15.0 M NH₃ (K_b = 1.8 × 10^-5) . Which of the following would be a dominant species in this solution?

Explain why we cannot directly compare K_sp values in comparing solubilities of ionic solids with different numbers of ions.