* OWLV2 | On X : OWLV2 | As: x OWLV2 | On Xy google - Yal xe Registration XD selfserve.un x/ilrn/takeAssignment/takeCovalentActivity.do?locator=assignment-takeUse the References to access important values if needed for this question.The molar solubility of iron(III) sulfide in a water solution isM.Submit AnswerRetry Entire Group8 more group attempts remainingChpinsort scdeletefgfi0>ト

For a salt AxBy, If molar solubility is S mol/l and solubility product is Ksp…

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The molar solubility of iron(III) sulfide in a water solution is M.

Fundamentals Of Analytical Chemistry

9th Edition

ISBN:9781285640686

Author:Skoog

Publisher:Skoog

Chapter10: Effect Of Electrolytes On Chemical Equilibria

Section: Chapter Questions

Problem 10.16QAP

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A 0.010 M solution of the weak acid HA has an osmotic pressure (see chapter on solutions and colloids) of 0.293 atm at 25 C. A 0.010 M solution of the weak acid HB has an osmotic pressure of 0.345 atm under the same conditions. (a) Which acid has the larger equilibrium constant for ionization HA[HA(aq)A(aq)+H+(aq)] or HB[HB(aq)H+(aq)+B(aq)]? (b) What are the equilibrium constants for the ionization of these acids? (Hint: Remember that each solution contains three dissolved species: the weak acid (HA or HB). the conjugate base (A- or B- and the hydrogen ion (H+). Remember that osmotic pressure (like all colligative properties) is related to the total number of solute particles. Specifically for osmotic pressure, those concentrations are described by molarities.)
Figure 16.3 shows the solubility of AgNO3 in water inunits of moles of AgNO3 per kilogram of H2O. If 255 g of AgNO3 is added to 100 g of water at 95°C and cooledslowly, at what temperature will the solution becomesaturated?
A solution is prepared by dissolving 40.00 g of NaCl (f.w. = 58.44 g mol–1), a non-volatile solute, in enough water (m.w. = 18.02 g mol–1) to result in exactly 1 L of solution at 25 °C. Assume the density of the solution is that of pure water (dsolution = 1.000 g mL–1). The ebullioscopic constant (Kb) for water is 0.513 °C m–1. The cryoscopic constant (Kf) for water is 1.86 °C m–1. The vapor pressure of pure water is 0.0313 atm. Find the freezing point of the solution(in C to 2 decimal places)
A solution is prepared by dissolving 40.00 g of NaCl (f.w. = 58.44 g mol–1), a non-volatile solute, in enough water (m.w. = 18.02 g mol–1) to result in exactly 1 L of solution at 25 °C. Assume the density of the solution is that of pure water (dsolution = 1.000 g mL–1). The ebullioscopic constant (Kb) for water is 0.513 °C m–1. The cryoscopic constant (Kf) for water is 1.86 °C m–1. The vapor pressure of pure water is 0.0313 atm. Find the vapor pressure of the solution to 3 decimal places in atm.
A solution is prepared by dissolving 40.00 g of NaCl (f.w. = 58.44 g mol–1), a non-volatile solute, in enough water (m.w. = 18.02 g mol–1) to result in exactly 1 L of solution at 25 °C. Assume the density of the solution is that of pure water (dsolution = 1.000 g mL–1). The ebullioscopic constant (Kb) for water is 0.513 °C m–1. The cryoscopic constant (Kf) for water is 1.86 °C m–1. The vapor pressure of pure water is 0.0313 atm. Find the osmotic pressure in atm to three decimal places
A solution is prepared by dissolving 40.00 g of NaCl (f.w. = 58.44 g mol–1), a non-volatile solute, in enough water (m.w. = 18.02 g mol–1) to result in exactly 1 L of solution at 25 °C. Assume the density of the solution is that of pure water (dsolution = 1.000 g mL–1). The ebullioscopic constant (Kb) for water is 0.513 °C m–1. The cryoscopic constant (Kf) for water is 1.86 °C m–1. The vapor pressure of pure water is 0.0313 atm. Determine the boiling point of the solution(in C to 2 decimal places)
A solution is prepared by dissolving 40.00 g of NaCl (f.w. = 58.44 g mol–1), a non-volatile solute, in enough water (m.w. = 18.02 g mol–1) to result in exactly 1 L of solution at 25 °C. Assume the density of the solution is that of pure water (dsolution = 1.000 g mL–1). The ebullioscopic constant (Kb) for water is 0.513 °C m–1. The cryoscopic constant (Kf) for water is 1.86 °C m–1. The vapor pressure of pure water is 0.0313 atm. Determine the following: Boiling point of solution (in °C to two decimal places) Freezing point of solution (in °C to two decimal places) Vapor pressure of the solution (in atm to three decimal places) Osmotic pressure (in atm to three decimal places)
1.0 GRAMS of an electrolyte was dissolved in 29 grams of water, the resulting solution was found to have a molarity of 0.437 mol / L. The freezing point of the solution was determined to be -4.96°C. The freezing point depression constant for water is -1.86°C and pure water may be assumed to freeze at 0°C. If the molecular weight of the electrolyte is 179.7 g/mol and it takes 15 minutes to perform the experiment, how much water was used (grams) ?
Solubility tests of benzoic acid with a dissolution enthalpy of -32kJ / mol were performed at 2 different temperatures. In the first experiment carried out at 27 ° C, 5 mL of the sample taken from the benzoic acid solution is taken into 25 mL of water and titrated with 0.02M NaOH. As a result of the titration, there is a consumption of 8 mL, respectively. In the second experiment performed at a different temperature under the same conditions, 22 mL was consumed. Taking advantage of these results a) Solubility of benzoic acid at 27 ° C .................................................; b) solubility of benzoic acid at different temperature ................................................; c) second test temperature ....................................................... (M (benzoic acid): 122.12 g / mol) Note: enter only required numerical values and units in the spaces in the question.
23. The Ksp of CuCO3ls) (copper carbonate, a salt) is reported as 2.21x10-11. Determine the solubility (in mg/L) of 1this salt in pure water Constants: The molecular mass (M) of copper carbonate (CuCOg) is 123.554 g mol-1
I would like to calculate the molar solubility and Ksp from the following data: I have calculated them (see the solutions in the table) but, I am concerned that I have done the calculation incorrectly. See images for my calculations. These calculations are for Sample 1. Thank you! Sample 1 Sample 2 Sample 3 Mass of Erlenmeyer Flask 23.59g 23.59g 23.60g Mass of Erlenmeyer Flask + Calcium Hydroxide Solution (lime water) 26.62g 26.65g 26.69g Mass of Calcium Hydroxide Solution 3.03g 3.06g 3.09g Volume of Ca(OH)2 Density = 1.000 g/mL 3.03mL 3.06mL 3.09mL Concentration of HCl (M) 0.1M 0.1M 0.1M Initial HCl Volume in Syringe 1.00mL(1) 1.00mL(2) 1.00mL(3) 1.00mL(1) 1.00mL(2) 1.00mL(3) 1.00mL(1) 1.00mL(2) 1.00mL(3) Final HCl Volume in Syringe 0ml (1) 0ml (2) 0.65mL (3) 0ml (1) 0ml (2) 0.63mL (3) 0ml (1) 0ml (2) 0.63mL (3) Volume of HCl Delivered 2.35mL 2.37mL 2.37mL Moles of HCl Delivered…
Which of the following compounds will be the MOST soluble in pure water at 25 ˚C? A. Lead (II) carbonate, PbCO3 Ksp = 7.4×10–14 B. Cobalt (II) carbonate, CoCO3 Ksp = 1.4×10–13 C. Manganese (II) carbonate, MnCO3 Ksp = 1.8×10–11 D. Cadmium carbonate, CdCO3 Ksp = 5.2×10–12 E. Barium carbonate, BaCO3 Ksp = 5.1×10–9