5.Why are some salts acidic when others are neutral?Curriculum Expectation E2. investigate the qualitative and quantitative nature of chemicalsystems at equilibrium, and solve related problems.1.Consider the following reaction:PCI (g)+ Cl2 (g) = PCI 5 (g)K = 4.30 x 10-6Initially, 2.50 M PCI, and 1.40 M Cl2 are placed into a container and allowed toreach equilibrium. Determine the equilibrium concentrations of all the species. 2.Would you expect a solution of ammonium chloride (NH_CI) to be acidic or basicwhen added to water? (K NH = 1.8 x 10-5)3.4.What is the pH of a 1.15 M solution of methanoic acid? K = 1.8 x 104What is the pH of a 0.350 M solution of lithium citrate? (K_ Citric Acid = 7.4 × 10-4)

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Answered: 5. Why are some salts acidic when…

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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Write balanced chemical equations that correspond to the following equilibrium constant expressions. (Use the lowest possible coefficients. Be sure to specify states such as (aq) or (s). If a box is not needed, leave it blank.) a. K b. K = = C. K = [H₂O₂] ² [H₂O]²[0₂] [Ag+] ² [SO₁²] [HSO3-] [H₂O+] [H₂SO3] TI स्टे TI + +
4. A solution of HF(aq) has an initial concentration of 0.75 mol/L. a) What is the pOH at equilibrium, if Ka = 7.1 x 104? b) What is the percent dissociation? Show your work.
14
MISSED THIS? Read Section 16.6 (Page); Watch IWE 16.5. Consider the following reaction: SO₂ Cl2 (g) SO2 (g) + Cl2 (g) A reaction mixture is made containing an initial [SO₂ C1₂] of 2.0×10-² M . At equilibrium, [Cl₂] = 1.1×10-² M. Part A Calculate the value of the equilibrium constant (Kc). Express your answer to two significant figures. for Part A for Part A do for Part redo fol Part A reset for Part. Ke= 2.2410-² Submit Previous Answers Request Answer X Incorrect; Try Again; 5 attempts remaining
3. At a certain temperature, Keq = 150 for the reaction being studied in today's lab. Use Equation (3) to calculate the concentrations of Fe³+ and SCN in a solution that has an equilibrium FeSCN²+ concentration of 4.5 x 105 M. Assume the equilibrium concentrations of Fe³+ and SCN are equal.
20. For each of the following equilibrium reactions and their associated K value, determine what species will be predominate at equilibrium.a. HC2H3O2(aq) + H2O(l) ⇄ H3O1+(aq) + C2H3O21-(aq) K=1.8x10-5b. Hydrochloric acid reacts with water to form the hydronium ion and the chloride ion. The K value is 1.8x1060c. Water reacts with water to form the hydronium ion and the hydroxide ion. The K value is 1.0x10-14d. NaC2H3O2(s) ⇄ Na1+(aq) + C2H3O21-(aq) K=2.5x1045
Experiment 3, Part 1: Write the equilibrium equation established in a saturated potassium chloride, KCl, solution. equilibrium reaction:
At 1,900 K, the equilibrium constant for the reaction below is Kc = 2.13. Br2(g) 2 Br(g) What is the percent dissociation of Br2 at 1,900 K, if the initial concentration of Br2 is 1.75mol L-1 ? Give your answer accurate to two significant figures. Do not include the % sign as part of your answer. The percent dissocation of Br2 is % ? Hints: If the initial concentration of Br2 is Co and the concentration of Br2 decreases by x(because of dissociation), then the percent dissociation of Br2 is equal to 100(x/ Co). You may have to use the quadratic formula to solve this problem. The quadratic formula, given below, can be used to find the solutions to the equation ax² + bx+c = 0.
Determine the pH of a solution of HCIO2 by constructing an ICE table, writing the equilibrium constant expression, and using this information to determine the pH. Complete Parts 1-3 before submitting your answer. 3 A 0.020 M aqueous solution of HCIO2 is prepared. Fill in the ICE table with the appropriate value for each involved species to determine concentrations of all reactants and products.. Initial (M) Change (M) Equilibrium (M) -2x 1.2 x 10-² - 2x 0 0.020 + x 1 HCIO₂(aq) 0.020 0.020 - x + 2 H₂O(l) 1.2 x 10-² 0.020 + 2x 0.010 0.020 - 2x H3O+ (aq) +X 1.2 x 10-² + x -X + NEXT 1.2 x 10-² - x CIO₂ (aq) RESET +2x 1.2 x 10-² + 2x >

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