1. Complete the reaction Mg(OH)2 Mg+2 +2ionization2. What is the pH of a 0.001000 M Mg(OH)2 solution ?first, find the [OH-1] using stoichiometry(1.0000 x 10-³ M Mg(OH)₂ )(-then, find the [H3O+¹] using KwKw[H3O+¹] =[pH = -log(mol OH-1mol Mg(OH)2d. perchloric[] [=b. hydrochlorich. hydrosulfuric0. H₂PO3-1 p. CO3-²-2c. phosphorici. OH-¹j. H30+1q. HCO3-1 r. H₂Sx. 1.000 x 10-³cc. 11.0000dd. 2v. 0.1000W. 0.02500bb. 1.000 x 10-11hh. 11.3010oo. baseii. 0.01250pp. conjugate basea. aceticg. sulfurous-2n. HPO3-²u. 0.2500aa. 2.6990gg. 5.000 x 10-12mm. conjugate acid-2k. SO4²jj. 0.22100e. hydrofluoric1. Mg+2there is completeМ ОН-1S. HS-1y. 3.0000ee. 1f. sulfurict. S-²z. 1x 10-14ff. 2.000 x 10-³kk. 0.05656 II. acidm. Cl-1

1. Complete the reaction Mg(OH)2 Mg+2 +2 ionization 2. What is the pH of a 0.001000 M Mg(OH)2 solution ? first, find the [OH-¹] using stoichiometry (1.0000 x 10-³ M Mg(OH)₂ )(- then, find the [H3O+¹] using Kw Kw [H3O+¹] = pH = -log( a. acetic b. hydrochloric g. sulfurous n. HPO3-2 0. H₂PO3-1 u. 0.2500 v. 0.1000 aa. 2.6990 gg. 5.000 x 10-12 mm. conjugate acid h. hydrosulfuric P. CO32 w. 0.02500 bb. 1.000 x 10-11 hh. 11.3010 oo. base [ D)= c. phosphoric i. OH-1 mol OH-1 = mol Mg(OH)2 d. perchloric j. H30+1 q. HCO3-1 r. H₂S x. 1.000 x 10-3 dd. 2 cc. 11.0000 ii. 0.01250 pp. conjugate base k. SO4² jj. 0.22100 e. hydrofluoric 1. Mg+2 S. HS-1 y. 3.0000 ee. 1 there is complete MOH-1 kk. 0.05656 t. S-² z. 1 x 10-14 ff. 2.000 x 10-3 f. sulfuric m. Cl-1 II. acid

1. Complete the reaction Mg(OH)2 Mg+2 +2 ionization 2. What is the pH of a 0.001000 M Mg(OH)2 solution ? first, find the [OH-¹] using stoichiometry (1.0000 x 10-³ M Mg(OH)₂ )(- then, find the [H3O+¹] using Kw Kw [H3O+¹] = pH = -log( a. acetic b. hydrochloric g. sulfurous n. HPO3-2 0. H₂PO3-1 u. 0.2500 v. 0.1000 aa. 2.6990 gg. 5.000 x 10-12 mm. conjugate acid h. hydrosulfuric P. CO32 w. 0.02500 bb. 1.000 x 10-11 hh. 11.3010 oo. base [ D)= c. phosphoric i. OH-1 mol OH-1 = mol Mg(OH)2 d. perchloric j. H30+1 q. HCO3-1 r. H₂S x. 1.000 x 10-3 dd. 2 cc. 11.0000 ii. 0.01250 pp. conjugate base k. SO4² jj. 0.22100 e. hydrofluoric 1. Mg+2 S. HS-1 y. 3.0000 ee. 1 there is complete MOH-1 kk. 0.05656 t. S-² z. 1 x 10-14 ff. 2.000 x 10-3 f. sulfuric m. Cl-1 II. acid

Chemistry: The Molecular Science

5th Edition

ISBN:9781285199047

Author:John W. Moore, Conrad L. Stanitski

Publisher:John W. Moore, Conrad L. Stanitski

Chapter14: Acids And Bases

Section: Chapter Questions

Problem 32QRT

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Similar questions

Define or illustrate the meaning of the following terms: a. amphoteric b. Kw reaction c. Kw equilibrium constant d. pH e. pOH f. pKw Give the conditions for a neutral aqueous solution at 25C, in terms of [H+], pH, and the relationship between [H+] and [OH]. Do the same for an acidic solution and for a basic solution. As a solution becomes more acidic, what happens to pH, pOH, [H+], and [OH]? As a solution becomes more basic, what happens to pH, pOH, [H+], and [OH]?
Two strategies are also followed when solving for the pH of a base in water. What is the strategy for calculating the pH of a strong base in water? List the strong bases mentioned in the text that should be committed to memory. Why is calculating the pH of Ca(OH)2 solutions a little more difficult than calculating the pH of NaOH solutions? Most bases are weak bases. The presence of what element most commonly results in basic properties for an organic compound? What is present on this element in compounds that allows it to accept a proton? Table 13-3 and Appendix 5 of the text list Kb values for some weak bases. What strategy is used to solve for the pH of a weak base in water? What assumptions are made when solving for the pH of weak base solutions? If the 5% rule fails, how do you calculate the pH of a weak base in water?
The ion HTe- is an amphiprotic species; it can act as either an acid or a base.. (a) What is Ka for the acid reaction of HTe- with H2O?. (b) What is Kb, for the reaction in which HTe- functions as a base in water?. (c) Demonstrate whether or not the second ionization of H2Te can be neglected in the calculation of [HTe-] in a 0.10 M solution of H2Te.
What is the pH of a solution obtained by mixing 235 mL of NaOH with a pH of 11.57 and 316 mL of Sr(OH)2 with a pH of 12.09? Assume that volumes are additive.
For conjugate acidbase pairs, how are Ka and Kb related? Consider the reaction of acetic acid in water CH3CO2H(aq)+H2O(l)CH3CO2(aq)+H3O+(aq) where Ka = 1.8 105 a. Which two bases are competing for the proton? b. Which is the stronger base? c. In light of your answer to part b. why do we classify the acetate ion (CH3CO2) as a weak base? Use an appropriate reaction to justify your answer. In general, as base strength increases, conjugate acid strength decreases. Explain why the conjugate acid of the weak base NH3 is a weak acid. To summarize, the conjugate base of a weak acid is a weak base and the conjugate acid of a weak base is a weak acid (weak gives you weak). Assuming Ka for a monoprotic strong acid is 1 106, calculate Kb for the conjugate base of this strong acid. Why do conjugate bases of strong acids have no basic properties in water? List the conjugate bases of the six common strong acids. To tie it all together, some instructors have students think of Li+, K+, Rb+, Cs+, Ca2+, Sr2+, and Ba2+ as the conjugate acids of the strong bases LiOH, KOH. RbOH, CsOH, Ca(OH)2, Sr(OH)2, and Ba(OH)2. Although not technically correct, the conjugate acid strength of these cations is similar to the conjugate base strength of the strong acids. That is, these cations have no acidic properties in water; similarly, the conjugate bases of strong acids have no basic properties (strong gives you worthless). Fill in the blanks with the correct response. The conjugate base of a weak acid is a_____base. The conjugate acid of a weak base is a_____acid. The conjugate base of a strong acid is a_____base. The conjugate acid of a strong base is a_____ acid. (Hint: Weak gives you weak and strong gives you worthless.)
What is the pH of a solution obtained by adding 13.0 g of NaOH to 795 mL of a 0.200 M solution of Sr(OH)2? Assume no volume change after NaOH is added.
At 25C, a 0.327 M solution of a weak acid HX has a pH of 5.12. What is G for the dissociation of the weak acid?
Does the pH of the solution increase, decrease, or stay the same when you (a) Add solid sodium oxalate, Na2C2O4, to 50.0 mL of 0.015-M oxalic acid? (b) Add solid ammonium chloride to 100. mL of 0.016-M HCl? (c) Add 20.0 g NaCl to 1.0 L of 0.012-M sodium acetate, NaCH3COO?
Estimate the pH that results when the following two solutions are mixed. a) 50 mL of 0.3 M CH3COOH and 50 mL of 0.4 M KOH b) 100 mL of 0.3 M CH3COOH and 50 mL of 0.4 M NaOH c) 150 mL of 0.3 M CH3COOH and 100 mL of 0.3 M Ba(OH)2 d) 200 mL of 0.3 M CH3COOH and 100 mL of 0.3 M Ba(OH)2
Composition diagrams, commonly known as alpha plots, are often used to visualize the species in a solution of an acid or base as the pH is varied. The diagram for 0.100 M acetic acid is shown here. The plot shows how the fraction [alpha ()] of acetic acid in solution, =[CH3CO2H][CH3CO2H]+[CH3CO2] changes as the pH increases (blue curve). (The red curve shows how the fraction of acetate ion, CH3CO2, changes as the pH increases.) Alpha plots are another way of viewing the relative concentrations of acetic acid and acetate ion as a strong base is added to a solution of acetic acid in the course of a titration. (a) Explain why the fraction of acetic acid declines and that of acetate ion increases as the pH increases. (b) Which species predominates at a pH of 4, acetic acid or acetate ion? What is the situation at a pH of 6? (c) Consider the point where the two lines cross. The fraction of acetic acid in the solution is 0.5, and so is that of acetate ion. That is, the solution is half acid and half conjugate base; their concentrations are equal. At this point, the graph shows the pH is 4.74. Explain why the pH at this point is 4 74.
Find [OH] and the pH of the following solutions. (a) 0.25 g barium hydroxide, Ba(OH)2, dissolved in enough water to make 0.655 L of solution (b) A 3.00 L solution of KOH is prepared by diluting 300.0 mL 0.149 M KOH with water.
1. Calculate the pH of 2.8 x 10^-4 M Ba(OH)2. 2. Calculate the pH of a 0.26 M methylamine solution. Kb = 4.4 x 10^-4 3.Calculate the pH of an aqueous solution containing the following: (a). 0.80 M lactic acid and 0.40 M lactate ion (Ka of lactic acid = 1.38 x 10^-4) (b). 0.30 M NH3 and 1.50 M NH4+ (Kb of NH3 = 1.8 x 10^-5)