= 1.6 x 10-10 Given the reactions below, calculate the molar solubility of AgCl (s) in 0.566 M NH3 (aq) at 25 degrees C. AgCl (s) →> Ag+ (aq) + Cl¯ (aq) K sp Ag+ (aq) + 2 NH3 (aq) →> Ag (NH3) + (aq) K = 1.7 x 107 f 2 Report your answer rounded to two significant figures and do NOT include units. Use "e" for scientific notation. For example, 1.0 x 10-14 is 1.0e-14.
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- The two salts BaCl2 and Ag2SO4 are both far more soluble in water than either BaSO4 (Ksp= 1.1 × 10-10) or AgCl (Ksp= 1.6 × 10-10) at 25°C. Suppose 50.0 mL of 0.040 M BaCl2(aq) is added to 50.0 mL of 0.020 M Ag2SO4(aq). Calculate the concentrations of SO42- (aq), Cl2(aq), Ba2+(aq), and Ag+(aq) that remain in solution at equilibrium.Calculate the equilibrium concentration of Ag*(aq) in a solution that is initially 0.130 M AgNO3 and 0.920 M NH3, and in which the following reaction takes place: Ag (aq) + 2NH₂ (aq) → Ag (NH₂)(ag) (K-1.70x107) 1.225 x 10-8 M AgCalculate the equilibrium concentration of Ag*(aq) in a solution that is initially 0.140 M AgNO3 and 0.980 MNH3, and in which the following reaction takes place: Ag*(aq) + 2NH3(aq)= Ag(NH3);(ag) (Kf = 1.70×107)
- What is the molar solubility (in mols/L)of silver chloride in 2.00 M NH3(aq) given that Kgp of AGCI = 1.77x10-10 and Kf of Ag(NH3)2* is 1.70×107? Express your answer in decimal notation rounded to three significant figures.Determine if the following salt is neutral, acidic or basic. If acidic or basic, write the appropriate equilibrium equation for the acid or base that exists when the salt is dissolved in aqueous solution. If neutral, simply write only NR. Be sure to include the proper phases for all species within the reaction. Ba(CHO) Ba(CHOz)z(aq) + 2 H2O(I)= Baz+(aq) + 2 CHO, (aq) + 2 OH-(aq)Write the equilibrium constant expression for acetic acid, and calculate a K value and an apparent percent dissociation for each concentration + ICE tables for 0.10 M HC2H302 +0.010 M +0.0010 M. (ph for 0.10M is 3.3, pH for 0.010 M is 3.7, ph for0.0010 M is 4.4) HA(aq) = H+(aq) + A-(aq)
- The weak monoprotic acid, acetic acid, is titrated with the strong base, potassium hydroxide as follows: HC2H3O2(aq) + K+ OH- (aq) → K+ C2H3O2-(aq) + H2O(l) Ka for acetic acid is 1.81 x 10-5 (at 25 oC). A 25.00 mL sample of a solution of acetic acid with concentration 0.0833 M is titrated with 0.1000 M KOH. a) Sketch the pH vs volume of added base titration curve for this reaction. pH vertically, volume of base horizontally. Label your axes with correct pH and volumes. (Attach more space, if needed) b) what is the pH at the beginning of the titration, Vbase = 0.00 mL ? c) what is the volume of the base needed to reach the equivalence point ? (label the equiv. point) d) what is the pH at the equivalence point?e) what is the pH of the titration when 5.00 mL of base have been added ? f) what is the pH when the volume of base added equals half the volume of the equivalence point? g) what is the pH of the titration when 20.00 mL of base have been added? h) what is the pH of the titration…The weak monoprotic acid, acetic acid, is titrated with the strong base, potassium hydroxide as follows: HC2H3O2(aq) + K+ OH- (aq) → K+ C2H3O2-(aq) + H2O(l) Ka for acetic acid is 1.81 x 10-5 (at 25 oC). A 25.00 mL sample of a solution of acetic acid with concentration 0.0833 M is titrated with 0.1000 M KOH. 1. what is the pH when the volume of base added equals half the volume of the equivalence point? 2. what is the pH of the titration when 20.00 mL of base have been added? 3. what is the pH of the titration when 30.00 mL of base have been added?One drop, 0.200 cm3, of 1.00 mol dm-3 of aqueous sodium hydroxide, NaOH, solution is added to 25.0 cm3 of a phosphate buffer that is 0.040 mol dm-3 in KH2PO4(aq) and 0.020 mol dm-3 in K2HPO4(aq). Calculate the resulting pH of the solution given that the second acid dissociation constant, pKa2, for phosphoric acid is 7.21.
- The weak monoprotic acid, acetic acid, is titrated with the strong base, potassium hydroxide as follows: HC2H3O2(aq) + K+ OH- (aq) → K+ C2H3O2-(aq) + H2O(l) Ka for acetic acid is 1.81 x 10-5 (at 25 oC). A 25.00 mL sample of a solution of acetic acid with concentration 0.0833 M is titrated with 0.1000 M KOH. A. What is the pH at the beginning of the titration, Vbase = 0.00 mL? B. What is the pH at the equivalence point? C. What is the pH of the titration when 5.00 mL of base have been added? D. What is the pH when the volume of base added equals half the volume of the equivalence point? E. What is the pH of the titration when 20.00 mL of base have been added? F. What is the pH of the titration when 30.00 mL of base have been added?One can calculate the molar solubility of a conic marble sample, predominantly made of CACO3, when exposed to acid rain with a pH of 4.34 using the following information: CaCO3(6) + H2O1) + CO2(aq) Ca2+, (aq) + 2 НСОЗ (аq) K = ? CaCO3(6) = Ca"(aq) + CO3“ (aq) Ksp = 2.8 x 109 CO2{aq) + 2 H2O) = H3O*(aq) + HCO3 (aq) K1 = 4.4 x 10-7 HCO3 (aq) + H2O) = H30*(aq) + CO3 (aq) K2 = 4.7 x 1011 2 H2O1) = H30* (aq) + OH (aq) Kw = 1.0 x 10-14 A water sample from around the marble is collected into a conical container. Assuming that the calcium is purely from the marble sample, and basing your calculations off of the molar solubility, calculate how many grams of calcium are present in this water sample. Note that the cone has a base radius of 10.00mm and a height of 480.8mm. molar solubility of CaCO3 = M amount of calcium x 10-3 g %3DWhat is the pH (aq., 25 oC) above which a solution of iron(III) hydroxide will precipitate? Ksp = 2.60 x 10–39.