1. Problem: Determine the reaction direction or equilibrium status for the reaction $$2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g)$$ with equilibrium constant $$K=0.33$$ and initial concentrations $$[SO_2]=0.39\,M$$, $$[O_2]=0.14\,M$$, $$[SO_3]=0.11\,M$$. 2. Calculate the reaction quotient $$Q$$: $$Q=\frac{[SO_3]^2}{[SO_2]^2[O_2]}=\frac{(0.11)^2}{(0.39)^2 \times 0.14} = \frac{0.0121}{0.0213} \approx 0.568$$ 3. Compare $$Q$$ to $$K$$: Since $$Q=0.568 > K=0.33$$, the reaction quotient is greater than the equilibrium constant. 4. Interpretation: When $$Q > K$$, the reaction will shift toward the reactants to reach equilibrium. 5. Final answer: The reaction will shift in the direction of reactants. --- 1. Problem: Calculate the concentration of $$Ag^+$$ ions in wastewater from the mass of $$Ag_2SO_4$$ precipitate. 2. Given: Mass of $$Ag_2SO_4 = 0.222\,g$$ Molar mass $$M = 311.80\,g/mol$$ Volume $$V = 34.38\,mL = 0.03438\,L$$ 3. Calculate moles of $$Ag_2SO_4$$: $$n = \frac{0.222}{311.80} = 0.000712\,mol$$ 4. Each mole of $$Ag_2SO_4$$ contains 2 moles of $$Ag^+$$ ions: $$n_{Ag^+} = 2 \times 0.000712 = 0.001424\,mol$$ 5. Calculate concentration of $$Ag^+$$: $$[Ag^+] = \frac{0.001424}{0.03438} = 0.0414\,mol/L$$ 6. Final answer: $$0.0414$$ --- 1. Problem: Calculate pH of 0.048 M $$Ca(OH)_2$$ solution. 2. $$Ca(OH)_2$$ dissociates completely: $$[OH^-] = 2 \times 0.048 = 0.096\,M$$ 3. Calculate pOH: $$pOH = -\log(0.096) = 1.02$$ 4. Calculate pH: $$pH = 14 - pOH = 14 - 1.02 = 12.98$$ 5. Final answer: $$12.98$$ --- 1. Problem: Calculate moles of $$SO_2$$ produced from reaction of 13.7 mol $$FeS_2$$ and 19.4 mol $$O_2$$: $$4FeS_2 + 11O_2 \rightarrow 2Fe_2O_3 + 8SO_2$$ 2. Calculate moles of $$SO_2$$ from limiting reactant. 3. Moles $$SO_2$$ from $$FeS_2$$: $$13.7\,mol\ FeS_2 \times \frac{8\,mol\ SO_2}{4\,mol\ FeS_2} = 27.4\,mol\ SO_2$$ 4. Moles $$SO_2$$ from $$O_2$$: $$19.4\,mol\ O_2 \times \frac{8\,mol\ SO_2}{11\,mol\ O_2} = 14.11\,mol\ SO_2$$ 5. Limiting reactant is $$O_2$$, so moles $$SO_2 = 14.11$$ 6. Final answer: $$14.1$$ (3 significant figures) --- 1. Problem: Identify net ionic equation for reaction of lithium phosphate and copper(II) nitrate. 2. Lithium phosphate dissociates as $$2Li_3PO_4 \rightarrow 6Li^+ + 2PO_4^{3-}$$ 3. Copper(II) nitrate dissociates as $$3Cu^{2+} + 6NO_3^-$$ 4. The precipitate formed is $$Cu_3(PO_4)_2(s)$$ 5. Net ionic equation: $$3Cu^{2+}(aq) + 2PO_4^{3-}(aq) \rightarrow Cu_3(PO_4)_2(s)$$ 6. Final answer: $$3Cu^{2+}(aq) + 2PO_4^{3-}(aq) \rightarrow Cu_3(PO_4)_2(s)$$ --- 1. Problem: Calculate pH of 0.87 M hypochlorous acid ($$HOCl$$) with $$K_a=3.5 \times 10^{-8}$$. 2. Set up expression: $$HOCl \rightleftharpoons H^+ + OCl^-$$ 3. Let $$x = [H^+]$$ at equilibrium. 4. $$K_a = \frac{x^2}{0.87 - x} \approx \frac{x^2}{0.87}$$ 5. Solve for $$x$$: $$x = \sqrt{K_a \times 0.87} = \sqrt{3.5 \times 10^{-8} \times 0.87} = 5.52 \times 10^{-5}$$ 6. Calculate pH: $$pH = -\log(5.52 \times 10^{-5}) = 4.26$$ 7. Final answer: $$4.26$$ --- 1. Problem: Identify acid with strongest conjugate base among $$HNO_2$$, $$CH_3COOH$$, $$HSO_4^-$$, $$HF$$, $$HClO$$. 2. Strongest conjugate base corresponds to weakest acid (lowest $$K_a$$). 3. Approximate $$K_a$$ values: - $$HNO_2$$: $$4.5 \times 10^{-4}$$ - $$CH_3COOH$$: $$1.8 \times 10^{-5}$$ - $$HSO_4^-$$: $$1.2 \times 10^{-2}$$ - $$HF$$: $$6.6 \times 10^{-4}$$ - $$HClO$$: $$3.0 \times 10^{-8}$$ 4. Weakest acid is $$HClO$$, so strongest conjugate base is $$ClO^-$$. 5. Final answer: $$HClO$$ --- 1. Problem: Calculate pH of 2.7 M methylamine ($$CH_3NH_2$$) with $$K_b=3.6 \times 10^{-4}$$. 2. Set $$x = [OH^-]$$ from base ionization. 3. $$K_b = \frac{x^2}{2.7 - x} \approx \frac{x^2}{2.7}$$ 4. Solve for $$x$$: $$x = \sqrt{K_b \times 2.7} = \sqrt{3.6 \times 10^{-4} \times 2.7} = 0.0312$$ 5. Calculate pOH: $$pOH = -\log(0.0312) = 1.51$$ 6. Calculate pH: $$pH = 14 - 1.51 = 12.49$$ 7. Final answer: $$12.49$$ --- 1. Problem: Calculate acid ionization constant $$K_a$$ for 0.75 M weak monoprotic acid ionized 94%. 2. Ionized concentration: $$[H^+] = 0.75 \times 0.94 = 0.705\,M$$ 3. Remaining acid concentration: $$0.75 - 0.705 = 0.045\,M$$ 4. Calculate $$K_a$$: $$K_a = \frac{[H^+][A^-]}{[HA]} = \frac{0.705 \times 0.705}{0.045} = 11.04$$ 5. Express in scientific notation: $$1.1 \times 10^1$$ 6. Final answer: $$1.1 \times 10^1$$ --- 1. Problem: Calculate molarity of weak base with $$K_b=3.36 \times 10^{-5}$$ and solution pH = 10.422. 2. Calculate pOH: $$pOH = 14 - 10.422 = 3.578$$ 3. Calculate $$[OH^-]$$: $$[OH^-] = 10^{-3.578} = 2.64 \times 10^{-4}$$ 4. Let $$c$$ be molarity of base, $$x = [OH^-]$$. 5. $$K_b = \frac{x^2}{c - x} \approx \frac{x^2}{c}$$ 6. Solve for $$c$$: $$c = \frac{x^2}{K_b} = \frac{(2.64 \times 10^{-4})^2}{3.36 \times 10^{-5}} = 0.00207$$ 7. Final answer: $$0.0021$$ --- 1. Problem: Identify reaction defined by $$pK_a$$ value. 2. $$pK_a$$ relates to acid dissociation: $$HF(aq) + H_2O(l) \rightleftharpoons F^-(aq) + H_3O^+(aq)$$ 3. Final answer: $$HF(aq) + H_2O(l) \rightleftharpoons F^-(aq) + H_3O^+(aq)$$ --- 1. Problem: Identify ion pair that cannot coexist in high concentrations. 2. Insoluble pairs include $$Ag^+$$ and $$Br^-$$ due to precipitation. 3. Final answer: $$Ag^+$$ and $$Br^-$$ --- 1. Problem: Identify Brønsted-Lowry acids in $$HCOO^-(aq) + H_2O(l) \rightleftharpoons HCOOH(aq) + OH^-(aq)$$ 2. Acids donate protons; here $$H_2O$$ donates proton to $$HCOO^-$$. 3. Acids are $$H_2O$$ and $$HCOOH$$. 4. Final answer: $$H_2O$$ and $$HCOOH$$ --- 1. Problem: Match compounds to pH boxes 1 (lowest) to 5 (highest) based on pKb or pKa. 2. Order of acidity/basicity: - Hydrogen sulfate ion (pKa=1.92) → box 1 (lowest pH) - Ammonia (pKb=4.74) → box 3 - Hypochlorite ion (pKb=6.54) → box 2 - Aniline (pKb=9.38) → box 4 - Phenol (pKa=9.89) → box 5 (highest pH) 3. Final answer: Hydrogen sulfate ion: 1 Hypochlorite ion: 2 Ammonia: 3 Aniline: 4 Phenol: 5 --- 1. Problem: Identify equilibria affected by volume change at constant temperature. 2. Volume affects equilibria with unequal moles of gas on each side. 3. Reactions: - 1: 2NO + 3F_2 ⇌ 2F_3NO (5 moles gas left, 5 right, no change) - 2: PCl_3 + Cl_2 ⇌ PCl_5 (2 moles left, 1 right, affected) - 3: O_3 + NO ⇌ NO_2 + O_2 (2 moles left, 2 right, no change) 4. Final answer: Only 2 --- 1. Problem: Identify salt giving acidic aqueous solution. 2. Cu(NO_3)_2 hydrolyzes to give acidic solution. 3. Final answer: $$Cu(NO_3)_2$$ --- 1. Problem: Calculate pH of pure water at temperature where $$K_w=5.95 \times 10^{-14}$$. 2. $$[H^+] = \sqrt{K_w} = \sqrt{5.95 \times 10^{-14}} = 7.71 \times 10^{-7}$$ 3. Calculate pH: $$pH = -\log(7.71 \times 10^{-7}) = 6.11$$ 4. Final answer: $$6.11$$ --- 1. Problem: Identify major species in 0.25 M $$NaNO_2$$ solution. 2. $$NaNO_2$$ dissociates into $$Na^+$$ and $$NO_2^-$$. 3. $$NO_2^-$$ is basic, so solution contains $$Na^+$$, $$NO_2^-$$, and $$OH^-$$. 4. Final answer: $$Na^+(aq), NO_2^-(aq), OH^-(aq)$$ --- 1. Problem: Calculate missing equilibrium constant $$K_2$$ for reaction: $$SO_3(g) \rightleftharpoons \frac{1}{2}O_2(g) + SO_2(g)$$ Given $$K_1 = 2.5 \times 10^6$$ for $$2SO_2 + O_2 \rightleftharpoons 2SO_3$$ 2. $$K_2 = \frac{1}{\sqrt{K_1}} = \frac{1}{(2.5 \times 10^6)^{0.5}} = 6.32 \times 10^{-4}$$ 3. Final answer: $$6.3 \times 10^{-4}$$ --- 1. Problem: Calculate pH of solution mixing 80.51 mL 0.040 M NaOH and 415.9 mL 0.020 M HCl. 2. Moles $$OH^-$$: $$0.040 \times 0.08051 = 0.0032204$$ 3. Moles $$H^+$$: $$0.020 \times 0.4159 = 0.008318$$ 4. Excess $$H^+$$ moles: $$0.008318 - 0.0032204 = 0.0050976$$ 5. Total volume: $$0.08051 + 0.4159 = 0.49641\,L$$ 6. $$[H^+] = \frac{0.0050976}{0.49641} = 0.01027$$ 7. Calculate pH: $$pH = -\log(0.01027) = 1.99$$ 8. Final answer: $$1.99$$