Calculating Equilibrium Concentrations - AP Chemistry

1. For the equilibrium $\text{COCl}_2(g) \rightleftharpoons \text{CO}(g) + \text{Cl}_2(g)$, $K_c = 8.0 \times 10^{-4}$ at 400°C. If the initial concentration of $\text{COCl}_2$ is 0.500 M and no products are initially present, what is the equilibrium concentration of $\text{CO}$?
   Answer: $0.0200 \text{ M}$ Let x = amount of COCl₂ that dissociates. At equilibrium: [COCl₂] = 0.500 - x, [CO] = [Cl₂] = x. Since Kc is small, assume x << 0.500. Then Kc = x²/(0.500) = 8.0 × 10⁻⁴. Solving: x² = 4.0 × 10⁻⁴, so x = 0.0200 M = [CO].
2. In a 1.00 L flask at constant temperature, the reaction $\mathrm{H_2(g) + I_2(g) \rightleftharpoons 2HI(g)}$ comes to equilibrium. Initially, $[\mathrm{H_2}] = 0.70\ \mathrm{M}$, $[\mathrm{I_2}] = 0.70\ \mathrm{M}$, and $[\mathrm{HI}] = 7.00\ \mathrm{M}$. Given $K_c = 49.0$, use an ICE table (assume only $[\mathrm{HI}]$ changes appreciably) to find the equilibrium concentration of $\mathrm{HI}$.
   Answer: 6.80 M This problem requires calculating equilibrium concentrations for H₂(g) + I₂(g) ⇌ 2HI(g). With initial concentrations [H₂] = 0.70 M, [I₂] = 0.70 M, and [HI] = 7.00 M, we calculate Q = [HI]²/([H₂][I₂]) = (7.00)²/(0.70)(0.70) = 100. Since Q > Kc (100 > 49), the reaction shifts left to reach equilibrium. Setting up the ICE table with [HI] decreasing by 2x: [H₂] = 0.70 + x, [I₂] = 0.70 + x, and [HI] = 7.00 - 2x at equilibrium. Substituting into Kc = 49: (7.00 - 2x)²/((0.70 + x)²) = 49, taking the square root gives (7.00 - 2x)/(0.70 + x) = 7. Solving: 7.00 - 2x = 4.90 + 7x, which gives x = 0.10, so [HI]eq = 7.00 - 2(0.10) = 6.80 M. The distractor 7.00 M (choice B) incorrectly assumes the system is already at equilibrium. The key strategy is to always check Q versus K and solve systematically using the equilibrium expression.
3. At equilibrium for $2\text{HI}(g) \rightleftharpoons \text{H}_2(g) + \text{I}_2(g)$, $[\text{HI}] = 0.80 \text{ M}$ and $[\text{H}_2] = [\text{I}_2] = 0.10 \text{ M}$. If 0.20 mol of additional HI is added to the 1.0 L container, what will be the new equilibrium concentration of $\text{H}_2$?
   Answer: $0.125 \text{ M}$ First calculate Kc from initial equilibrium: Kc = [H₂][I₂]/[HI]² = (0.10)(0.10)/(0.80)² = 0.0156. After adding 0.20 mol HI to 1.0 L: [HI] = 1.00 M, [H₂] = [I₂] = 0.10 M. Let x = mol/L of H₂ and I₂ formed. At new equilibrium: [HI] = 1.00 - 2x, [H₂] = [I₂] = 0.10 + x. Substituting into Kc: (0.10 + x)²/(1.00 - 2x)² = 0.0156. Taking square root: (0.10 + x)/(1.00 - 2x) = 0.125. Solving: 0.10 + x = 0.125 - 0.25x, so 1.25x = 0.025, giving x = 0.020 M. Therefore [H₂] = 0.10 + 0.025 = 0.125 M.
4. The reaction $N_2O_4(g) \rightleftharpoons 2NO_2(g)$ has an equilibrium constant $K_p = 0.66$. If a container is initially filled with only $NO_2$ at a pressure of 1.0 atm, what is the partial pressure of $N_2O_4$ at equilibrium?
   Answer: $0.28~atm$ Since only product is present initially, the reaction will proceed in reverse. Let x be the equilibrium partial pressure of $N_2O_4$. The change in $NO_2$ pressure will be -2x. At equilibrium, $P_{N_2O_4} = x$ and $P_{NO_2} = 1.0 - 2x$. The equilibrium expression is $K_p = \frac{(P_{NO_2})^2}{P_{N_2O_4}}$. So, $0.66 = \frac{(1.0 - 2x)^2}{x}$. This gives the quadratic equation $4x^2 - 4.66x + 1.0 = 0$. Solving for x gives two possible values, but only $x \approx 0.28$ atm results in a positive pressure for $NO_2$. Thus, the equilibrium pressure of $N_2O_4$ is $0.28$ atm.
5. At 500 K, the equilibrium constant $K_c = 0.040$ for the reaction $\text{PCl}_5(g) \rightleftharpoons \text{PCl}_3(g) + \text{Cl}_2(g)$. If 2.0 mol of $\text{PCl}_5$ is placed in a 4.0 L container and allowed to reach equilibrium, what is the equilibrium concentration of $\text{PCl}_3$?
   Answer: $0.12 \text{ M}$ Initial [PCl₅] = 2.0 mol/4.0 L = 0.50 M. Let x = amount that dissociates. At equilibrium: [PCl₅] = 0.50 - x, [PCl₃] = [Cl₂] = x. Kc = x²/(0.50 - x) = 0.040. Solving: x² + 0.040x - 0.020 = 0. Using the quadratic formula: x = 0.12 M.
6. A student prepares a 1.00 L mixture for the reversible reaction $\mathrm{H_2(g) + I_2(g) \rightleftharpoons 2HI(g)}$ at constant temperature. Initially, $[\mathrm{H_2}] = 0.20\ \mathrm{M}$, $[\mathrm{I_2}] = 0.20\ \mathrm{M}$, and $[\mathrm{HI}] = 4.00\ \mathrm{M}$. Given $K_c = 100$, and using an ICE table with the assumption that only $[\mathrm{HI}]$ changes appreciably, what is $[\mathrm{HI}]$ at equilibrium?
   Answer: 3.80 M This question asks for equilibrium concentration calculations using an ICE table approach. For H₂(g) + I₂(g) ⇌ 2HI(g) with initial concentrations [H₂] = 0.20 M, [I₂] = 0.20 M, and [HI] = 4.00 M, we first calculate Q = [HI]²/([H₂][I₂]) = (4.00)²/(0.20)(0.20) = 400. Since Q > Kc (400 > 100), the reaction shifts left to reach equilibrium. Setting up the ICE table with [HI] decreasing by 2x gives: [H₂] = 0.20 + x, [I₂] = 0.20 + x, and [HI] = 4.00 - 2x at equilibrium. Substituting into Kc = 100: (4.00 - 2x)²/((0.20 + x)²) = 100, taking the square root yields (4.00 - 2x)/(0.20 + x) = 10. Solving: 4.00 - 2x = 2.00 + 10x, which gives x = 0.10, so [HI]eq = 4.00 - 2(0.10) = 3.80 M. The distractor 4.00 M (choice C) incorrectly assumes no change from initial conditions. The key is recognizing that even small initial reactant concentrations require the system to shift when Q ≠ K.
7. Initially, 4.0 moles of gaseous reactant A are placed in an empty 2.0 L flask and allowed to establish equilibrium according to the reaction $2A(g) \rightleftharpoons B(g) + C(g)$. The equilibrium constant, $K_c$, is 0.25 for this reaction. What is the equilibrium concentration of B?
   Answer: $0.50 M$ First, calculate the initial concentration of A: $[A]_{initial} = \frac{4.0 \text{ mol}}{2.0 \text{ L}} = 2.0 M$. Let x be the equilibrium concentration of B. Then $[C] = x$ and $[A] = 2.0 - 2x$. The equilibrium expression is $K_c = \frac{[B][C]}{[A]^2} = \frac{(x)(x)}{(2.0 - 2x)^2} = 0.25$. Taking the square root of both sides gives $\frac{x}{2.0 - 2x} = \sqrt{0.25} = 0.50$. Solving for x: $x = 0.50(2.0 - 2x) = 1.0 - x$. This gives $2x = 1.0$, so $x = 0.50 M$. The equilibrium concentration of B is $0.50 M$.
8. The equilibrium $\text{Fe}^{3+}(aq) + \text{SCN}^-(aq) \rightleftharpoons \text{FeSCN}^{2+}(aq)$ has $K_c = 890$ at 25°C. If initial concentrations are $[\text{Fe}^{3+}] = 1.0 \times 10^{-3} \text{ M}$ and $[\text{SCN}^-] = 2.0 \times 10^{-4} \text{ M}$, what is the equilibrium concentration of $\text{FeSCN}^{2+}$?
   Answer: $1.8 \times 10^{-4} \text{ M}$ Let x = [FeSCN²⁺] formed. At equilibrium: [Fe³⁺] = 1.0 × 10⁻³ - x, [SCN⁻] = 2.0 × 10⁻⁴ - x, [FeSCN²⁺] = x. Since Kc is large, assume reaction goes nearly to completion with SCN⁻ limiting. Setting up: Kc = x/[(1.0 × 10⁻³ - x)(2.0 × 10⁻⁴ - x)] = 890. Assuming x ≈ 2.0 × 10⁻⁴, then [Fe³⁺] ≈ 8.0 × 10⁻⁴ and [SCN⁻] ≈ 0. Solving iteratively gives x = 1.8 × 10⁻⁴ M.
9. In a 1.00 L flask at a constant temperature, the reversible reaction $\mathrm{H_2(g) + I_2(g) \rightleftharpoons 2HI(g)}$ is established. Initially, $[\mathrm{H_2}] = 1.00\ \mathrm{M}$, $[\mathrm{I_2}] = 1.00\ \mathrm{M}$, and $[\mathrm{HI}] = 10.0\ \mathrm{M}$. The equilibrium constant is $K_c = 100$. Using an ICE table (assume only $[\mathrm{HI}]$ changes appreciably), what is the equilibrium concentration of $\mathrm{HI}$?
   Answer: 9.50 M This problem requires calculating equilibrium concentrations using an ICE table. For the reaction H₂(g) + I₂(g) ⇌ 2HI(g), we set up the ICE table with initial concentrations [H₂] = 1.00 M, [I₂] = 1.00 M, and [HI] = 10.0 M. Since we're told only [HI] changes appreciably, we let the change in [HI] be -2x (negative because Q > K, so the reaction shifts left), making changes in [H₂] and [I₂] each +x. At equilibrium: [H₂] = 1.00 + x, [I₂] = 1.00 + x, and [HI] = 10.0 - 2x. Substituting into Kc = [HI]²/([H₂][I₂]) = 100 gives (10.0 - 2x)²/((1.00 + x)(1.00 + x)) = 100, which simplifies to (10.0 - 2x)² = 100(1.00 + x)². Taking the square root: 10.0 - 2x = 10(1.00 + x), solving gives x = 0.25, so [HI]eq = 10.0 - 2(0.25) = 9.50 M. A common error (choice B, 10.0 M) assumes no change occurs because the initial [HI] seems large. The key strategy is to properly set up the ICE table with correct stoichiometric coefficients and solve the resulting equation systematically.
10. In a 1.00 L container at constant temperature, the equilibrium $\mathrm{H_2(g) + I_2(g) \rightleftharpoons 2HI(g)}$ is established. Initially, $[\mathrm{H_2}] = 0.80\ \mathrm{M}$, $[\mathrm{I_2}] = 0.80\ \mathrm{M}$, and $[\mathrm{HI}] = 8.00\ \mathrm{M}$. The value of $K_c$ is 64.0. Using an ICE table (assume only $[\mathrm{HI}]$ changes appreciably), what is the equilibrium concentration of $\mathrm{HI}$?
    Answer: 7.50 M This problem requires calculating equilibrium concentrations for the reaction H₂(g) + I₂(g) ⇌ 2HI(g). With initial concentrations [H₂] = 0.80 M, [I₂] = 0.80 M, and [HI] = 8.00 M, we calculate Q = [HI]²/([H₂][I₂]) = (8.00)²/(0.80)(0.80) = 100. Since Q > Kc (100 > 64), the reaction shifts left. Using an ICE table where [HI] changes by -2x: [H₂] = 0.80 + x, [I₂] = 0.80 + x, and [HI] = 8.00 - 2x at equilibrium. Substituting into Kc = 64: (8.00 - 2x)²/((0.80 + x)²) = 64, taking the square root gives (8.00 - 2x)/(0.80 + x) = 8. Solving: 8.00 - 2x = 6.40 + 8x, which yields x = 0.25, so [HI]eq = 8.00 - 2(0.25) = 7.50 M. The distractor 8.00 M (choice D) incorrectly assumes the system is already at equilibrium. The strategy involves calculating Q, determining shift direction, and solving the equilibrium expression with proper stoichiometry.
11. At 2000 K, $K_c = 4.0 \times 10^{-4}$ for $\text{N}_2(g) + \text{O}_2(g) \rightleftharpoons 2\text{NO}(g)$. If air containing $P_{\text{N}_2} = 0.78 \text{ atm}$ and $P_{\text{O}_2} = 0.21 \text{ atm}$ is heated to 2000 K, what is the equilibrium partial pressure of NO?
    Answer: $8.1 \times 10^{-3} \text{ atm}$ Convert to Kp: Kp = Kc(RT)^Δn where Δn = 2-1-1 = 0, so Kp = Kc = 4.0 × 10⁻⁴. Let 2x = amount of NO formed. At equilibrium: PN₂ = 0.78 - x, PO₂ = 0.21 - x, PNO = 2x. Since Kp is small, assume x << 0.21. Then Kp = (2x)²/[(0.78)(0.21)] = 4.0 × 10⁻⁴. Solving: 4x² = 4.0 × 10⁻⁴ × 0.164 = 6.56 × 10⁻⁵, so x² = 1.64 × 10⁻⁵, giving x = 4.05 × 10⁻³. Therefore PNO = 2x = 8.1 × 10⁻³ atm.
12. The equilibrium $\mathrm{H_2(g) + Cl_2(g) \rightleftharpoons 2HCl(g)}$ is established in a 1.0 L flask. Initially, $[\mathrm{H_2}]=0.010\,\mathrm{M}$, $[\mathrm{Cl_2}]=1.00\,\mathrm{M}$, and $[\mathrm{HCl}]=0\,\mathrm{M}$. At this temperature, $K_c=9.0\times 10^{-4}$. Using an ICE table and treating $[\mathrm{Cl_2}]$ as effectively constant, what is the equilibrium concentration of $\mathrm{HCl}$?
    Answer: 3.0×10⁻³ M This problem requires calculating equilibrium concentrations for H₂(g) + Cl₂(g) ⇌ 2HCl(g). Setting up an ICE table with initial values [H₂]=0.010 M, [Cl₂]=1.00 M, and [HCl]=0 M, we can treat [Cl₂] as constant due to its large initial concentration. The equilibrium expression is Kc = [HCl]²/([H₂][Cl₂]) = 9.0×10⁻⁴. If x mol/L of H₂ reacts, then [HCl]=2x at equilibrium and [H₂]=0.010-x. For small x, the expression becomes (2x)²/(0.010×1.00) = 9.0×10⁻⁴, which gives 4x²/0.010 = 9.0×10⁻⁴, so x² = 2.25×10⁻⁶, and x = 1.5×10⁻³ M. Therefore, [HCl] = 2x = 3.0×10⁻³ M. A common error would be to use [HCl]=x instead of [HCl]=2x in the ICE table, leading to an incorrect answer of 1.5×10⁻³ M. The key strategy is to carefully track stoichiometric coefficients when setting up the ICE table and ensure the equilibrium expression reflects the balanced equation.
13. Consider the reversible reaction: $N_2(g) + O_2(g) \rightleftharpoons 2NO(g)$. Initially, a 1.0 L flask contains 0.50 mol of $N_2$ and 0.50 mol of $O_2$. If the equilibrium constant, $K_c$, for the reaction is 0.10 at the temperature of the experiment, what is the equilibrium concentration of NO?
    Answer: $0.14 M$ Let x be the change in concentration of $N_2$ and $O_2$. The initial concentrations are $[N_2] = 0.50 M$ and $[O_2] = 0.50 M$. At equilibrium, $[N_2] = 0.50 - x$, $[O_2] = 0.50 - x$, and $[NO] = 2x$. The equilibrium expression is $K_c = \frac{[NO]^2}{[N_2][O_2]}$. Substituting the values: $0.10 = \frac{(2x)^2}{(0.50 - x)(0.50 - x)} = \frac{(2x)^2}{(0.50 - x)^2}$. Taking the square root of both sides gives $\sqrt{0.10} = \frac{2x}{0.50 - x}$, so $0.316 = \frac{2x}{0.50 - x}$. Solving for x gives $0.158 - 0.316x = 2x$, which simplifies to $0.158 = 2.316x$, so $x \approx 0.068 M$. The equilibrium concentration of NO is $2x$, which is $2 \times 0.068 M = 0.136 M$, or approximately $0.14 M$.
14. In a 1.00 L flask at a constant temperature, the reaction $\mathrm{H_2(g) + I_2(g) \rightleftharpoons 2HI(g)}$ reaches equilibrium. Initially, $[\mathrm{H_2}] = 0.50\ \mathrm{M}$, $[\mathrm{I_2}] = 0.50\ \mathrm{M}$, and $[\mathrm{HI}] = 5.00\ \mathrm{M}$. The equilibrium constant is $K_c = 25.0$. Using an ICE table (assume only $[\mathrm{HI}]$ changes appreciably), what is the equilibrium concentration of $\mathrm{HI}$?
    Answer: 4.50 M This problem involves calculating equilibrium concentrations for H₂(g) + I₂(g) ⇌ 2HI(g). Setting up the ICE table with initial values [H₂] = 0.50 M, [I₂] = 0.50 M, and [HI] = 5.00 M, we need to determine which direction the reaction shifts. Calculating Q = [HI]²/([H₂][I₂]) = (5.00)²/(0.50)(0.50) = 100, and since Q > Kc (100 > 25), the reaction shifts left. With [HI] changing by -2x, we have [H₂] = 0.50 + x, [I₂] = 0.50 + x, and [HI] = 5.00 - 2x at equilibrium. Substituting into Kc = 25: (5.00 - 2x)²/((0.50 + x)²) = 25, taking the square root gives (5.00 - 2x)/(0.50 + x) = 5. Solving: 5.00 - 2x = 2.50 + 5x, which gives x = 0.25, so [HI]eq = 5.00 - 2(0.25) = 4.50 M. A tempting distractor (choice D, 5.00 M) incorrectly assumes the system is already at equilibrium. The strategy is to always calculate Q first, compare to K to determine shift direction, then solve the equilibrium expression.
15. In a 1.00 L flask at constant temperature, the equilibrium $\mathrm{H_2(g) + I_2(g) \rightleftharpoons 2HI(g)}$ is established. Initially, $[\mathrm{H_2}] = 0.25\ \mathrm{M}$, $[\mathrm{I_2}] = 0.25\ \mathrm{M}$, and $[\mathrm{HI}] = 5.00\ \mathrm{M}$. If $K_c = 100$, and an ICE table is used with the assumption that only $[\mathrm{HI}]$ changes appreciably, what is the equilibrium concentration of $\mathrm{HI}$?
    Answer: 4.75 M This problem involves calculating equilibrium concentrations for H₂(g) + I₂(g) ⇌ 2HI(g). Starting with [H₂] = 0.25 M, [I₂] = 0.25 M, and [HI] = 5.00 M, we calculate Q = [HI]²/([H₂][I₂]) = (5.00)²/(0.25)(0.25) = 400. Since Q > Kc (400 > 100), the reaction must shift left. Setting up the ICE table with [HI] changing by -2x: [H₂] = 0.25 + x, [I₂] = 0.25 + x, and [HI] = 5.00 - 2x at equilibrium. Substituting into Kc = 100: (5.00 - 2x)²/((0.25 + x)²) = 100, taking the square root gives (5.00 - 2x)/(0.25 + x) = 10. Solving: 5.00 - 2x = 2.50 + 10x, which gives x = 0.125, so [HI]eq = 5.00 - 2(0.125) = 4.75 M. A common mistake (choice D, 5.00 M) assumes the initial concentration doesn't change. The key is recognizing that Q > K requires a leftward shift, even with seemingly balanced initial concentrations.
16. For the reaction $\text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g)$, the initial concentrations are $[\text{N}_2] = 0.50 \text{ M}$, $[\text{H}_2] = 1.50 \text{ M}$, and $[\text{NH}_3] = 0 \text{ M}$. At equilibrium, $[\text{NH}_3] = 0.20 \text{ M}$. What is the equilibrium concentration of $\text{N}_2$?
    Answer: $0.40 \text{ M}$ Using an ICE table: Initial [N₂] = 0.50 M, change = -x, equilibrium = 0.50 - x. Since 2 mol NH₃ are formed from 1 mol N₂, and [NH₃] at equilibrium = 0.20 M, then x = 0.20/2 = 0.10 M. Therefore, [N₂] at equilibrium = 0.50 - 0.10 = 0.40 M.
17. At 1000 K, $K_c = 0.263$ for $\text{C}(s) + 2\text{H}_2(g) \rightleftharpoons \text{CH}_4(g)$. If 2.00 mol of $\text{H}_2$ is placed with excess carbon in a 2.00 L container, what is the equilibrium concentration of $\text{CH}_4$?
    Answer: $0.265 \text{ M}$ Initial [H₂] = 2.00 mol/2.00 L = 1.00 M. Let x = [CH₄] formed. At equilibrium: [H₂] = 1.00 - 2x, [CH₄] = x. Kc = x/(1.00 - 2x)² = 0.263. Solving the quadratic equation: 1.052x² - 1.053x + 0.263 = 0 gives x = 0.265 M.
18. For the reaction $\text{I}_2(g) \rightleftharpoons 2\text{I}(g)$, the equilibrium constant $K_c$ is $3.8 \times 10^{-5}$. If the initial concentration of $\text{I}_2$ is 0.050 M, what is the approximate equilibrium concentration of I?
    Answer: $1.4 \times 10^{-3}\ \text{M}$ Let x be the change in concentration of $I_2$. At equilibrium, $[I_2] = 0.050 - x$ and $[I] = 2x$. The equilibrium expression is $K_c = \frac{[I]^2}{[I_2]}$. So, $3.8 \times 10^{-5} = \frac{(2x)^2}{0.050 - x}$. Since $K_c$ is small, we can approximate $0.050 - x \approx 0.050$. The equation becomes $3.8 \times 10^{-5} \approx \frac{4x^2}{0.050}$. Solving for $x^2$ gives $x^2 = \frac{(3.8 \times 10^{-5})(0.050)}{4} = 4.75 \times 10^{-7}$. So, $x \approx 6.9 \times 10^{-4} M$. The equilibrium concentration of I is $2x$, which is $2 \times (6.9 \times 10^{-4}) \approx 1.38 \times 10^{-3} M$, or approximately $1.4 \times 10^{-3} M$.
19. At 427°C, $K_c = 55.3$ for $\text{H}_2(g) + \text{Br}_2(g) \rightleftharpoons 2\text{HBr}(g)$. If the initial concentrations are $[\text{H}_2] = 0.200 \text{ M}$, $[\text{Br}_2] = 0.100 \text{ M}$, and $[\text{HBr}] = 0.300 \text{ M}$, what is the equilibrium concentration of $\text{Br}_2$?
    Answer: $0.0565 \text{ M}$ Calculate Q: Q = [HBr]²/([H₂][Br₂]) = (0.300)²/(0.200 × 0.100) = 0.09/0.02 = 4.5. Since Q < Kc (55.3), reaction proceeds forward. Let x = mol/L of H₂ and Br₂ consumed. At equilibrium: [H₂] = 0.200 - x, [Br₂] = 0.100 - x, [HBr] = 0.300 + 2x. Substituting: Kc = (0.300 + 2x)²/[(0.200 - x)(0.100 - x)] = 55.3. Expanding: (0.09 + 1.2x + 4x²)/[(0.02 - 0.3x + x²)] = 55.3. Solving the resulting quadratic gives x = 0.0435 M, so [Br₂] = 0.100 - 0.0435 = 0.0565 M.
20. The equilibrium $\text{Ag}^+(aq) + 2\text{NH}_3(aq) \rightleftharpoons [\text{Ag}(\text{NH}_3)_2]^+(aq)$ has $K_f = 1.6 \times 10^7$. If $[\text{Ag}^+]_0 = 0.010 \text{ M}$ and $[\text{NH}_3]_0 = 0.20 \text{ M}$, what is the equilibrium concentration of free $\text{Ag}^+$?
    Answer: $3.5 \times 10^{-6} \text{ M}$ Since Kf is very large, assume complete complex formation. Initially: [Ag(NH₃)₂⁺] ≈ 0.010 M, [NH₃] ≈ 0.18 M. Let x = [Ag⁺] at equilibrium. Then Kf = 0.010/[x(0.18)²] = 1.6 × 10⁷. Solving: x = 0.010/(1.6 × 10⁷ × 0.0324) = 3.5 × 10⁻⁶ M.

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