**Determining the Solubility Product Constant (Ksp) for Mg(CN)₂**The molar solubility of Mg(CN)₂ is given as 1.4 × 10⁻⁵ M at a certain temperature. Our objective is to determine the value of the solubility product constant (Ksp) for Mg(CN)₂.**Guidance:**Fill in the ICE (Initial, Change, Equilibrium) table to determine the concentrations of reactants and products in equilibrium based on the given solubility.### ICE Table| | Mg(CN)₂(s) | ⇌ | Mg²⁺(aq) | + | 2 CN⁻(aq) ||----------------------|------------|---|----------|---|-----------|| **Initial (M)** | | | | | || **Change (M)** | | | | | || **Equilibrium (M)** | | | | | |### Choices for ValuesBelow the table, several options are available for each entry in the table:- **Initial (M):** Usually starts at 0 for product ions as the solid begins to dissolve.- **Change (M):** Indicated by typical chemical equilibrium terms such as ±x, 1.4 × 10⁻⁵, or 1.4 × 10⁻⁵ ± x.- **Equilibrium (M):** Final concentrations are computed based on the changes that occurred.Utilize these values to adequately fill the table and compute Ksp using the relation:\[ K_{sp} = [\text{Mg}^{2+}][\text{CN}^-]^2 \]This setup guides the calculations required to find the equilibrium concentrations and support the derivation of Ksp. **Question 19 of 25**The molar solubility of Mg(CN)₂ is \(1.4 \times 10^{-5}\) M at a certain temperature. Determine the value of Ksp for Mg(CN)₂.**Instructions:**Based on the setup of your ICE table, construct the expression for Ksp and then evaluate it. Do not combine or simplify terms.\[ K_{sp} = \, \text{[ ]} = \, \text{[ ]} \]**Available Options:**- \([0]\)- \([1.4 \times 10^{-5}]\)- \([2.8 \times 10^{-5}]\)- \([1.4 \times 10^{-5}]^2\)- \([2.8 \times 10^{-5}]^2\)- \([x]\)- \([2x]^2\)- \([1.4 \times 10^{-5} + x]\)- \([1.4 \times 10^{-5} - x]\)- \([1.4 \times 10^{-5} - 2x]\)- \([1.4 \times 10^{-5} + x]^2\)- \([2.8 \times 10^{-5} - x]\)- \([2.8 \times 10^{-5} + 2x]\)**Expression Options:**- \(2.7 \times 10^{-15}\)- \(1.1 \times 10^{-14}\)- \(2.2 \times 10^{-14}\)- \(3.9 \times 10^{-10}\)**Reset Button:**To start over, press the "RESET" button.

Answered: Image /qna-images/answer/b9bcd809-f401-45a8-bde7-f8da10a65cdd.jpg

Question 19 of 25 The molar solubility of Mg(CN)2 is 1.4 x 10-5 M at a certain temperature. Determine the value of Ksp for Mg(CN). NEXT Based on the given values, fill in the ICE table to determine concentrations of all reactants and products. Mg(CN)-(s) Mg (aq) 2 CN-(aq) Initial (M) Change (M) Equilibrium (M) O RESET 1.4 x 105 -1.4 x 10-5 2.8 x 10-5 -2.8 x 10-5 +x +2x -2x 1.4 x 10-5 +x 1.4 x 10- + 2x 1.4 x 10-5 - x 1.4 x 10- - 2x 2.8 x 10-5 + x 2.8 x 10+ 2r -X 28x 10-5 -x 2.8 x 10- 2x

Question 19 of 25 The molar solubility of Mg(CN)2 is 1.4 x 10-5 M at a certain temperature. Determine the value of Ksp for Mg(CN). NEXT Based on the given values, fill in the ICE table to determine concentrations of all reactants and products. Mg(CN)-(s) Mg (aq) 2 CN-(aq) Initial (M) Change (M) Equilibrium (M) O RESET 1.4 x 105 -1.4 x 10-5 2.8 x 10-5 -2.8 x 10-5 +x +2x -2x 1.4 x 10-5 +x 1.4 x 10- + 2x 1.4 x 10-5 - x 1.4 x 10- - 2x 2.8 x 10-5 + x 2.8 x 10+ 2r -X 28x 10-5 -x 2.8 x 10- 2x

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...

See similar textbooks

Similar questions

An unknown metal, M, makes a precipitate with sulfate having the chemical formula M SO. A At equilibrium, the concentration of the metal ion is 0.000338 M and the sulfate ion concentration is 0.000169 M. What is the K of the metal sulfate? sp K = sp
An unknown metal, M, makes a precipitate with sulfate having the chemical formula M2SO4. At equilibrium, the concentration of the metal ion is 0.000889 M and the sulfate ion concentration is 0.0004445 M. What is the Ksp of the metal sulfate? Ksp =
What MASS of carbon dioxide would be dissolved in an Olympic size swimming pool (2.5 x 106 L) that is at equilibrium with air at 25˚C? The average partial pressure of carbon dioxide is currently 4.04 x 10-4 atm. Henry’s constant for carbon dioxide at this temperature is 3.4 x 10-2 M/atm. CO2 = 44.01 g/mol.
Suppose a 500 ml flask is filled with 0.20 mol of OCI,, 0.90 mol of BrOCI and 0.10 mol of BrCl. The following reaction becomes possible: Br₂(g) + OCI, (g) BrOCI(g) + BrC1 (g) The equilibrium constant K for this reaction is 7.85 at the temperature of the flask. Calculate the equilibrium molarity of OCI₂. Round your answer to two decimal places.
A sample of pure NO2 is heated to 339 °C at which temperature it partially dissociates according to the equation 2 NO₂ (g) ⇒ 2 NO (g) + O2 (g) At equilibrium the density of the gas mixture is 0.520 g/L at 0.750 atm.
If 0.54 molmol of Br2Br2 and 1.23 molmol of Cl2Cl2 are introduced into a 3.0-LL container at 400 KK, what will be the equilibrium concentration of Br2Br2? If 0.54 molmol of Br2Br2 and 1.23 molmol of Cl2Cl2 are introduced into a 3.0-LL container at 400 KK, what will be the equilibrium concentration of Cl2Cl2? If 0.54 molmol of Br2Br2 and 1.23 molmol of Cl2Cl2 are introduced into a 3.0-LL container at 400 KK, what will be the equilibrium concentration of BrClBrCl?
What is the equillibrium constant for COCl2(g) -- CO(g) + Cl2(g) if COCl2 concentration is 0.5 M and CO concentration is 1.34 x 10 -4 M at equillibrium?
i have a reaction of 2NO2(g) + Br2(g) <----> 2NOBr (g) Keq = 1.3 x 10 ^-2, temp = 1000 K how do I figure out the (new) Keq for the reverse and increased concentration reaction of: 4NOBr(g) <----> 4NO2 (g) + 2Br2(g)
) Listen The Kep value for calcium sulfate [CaSO4) is 2.40 X 10-5 and a professor made 1825 mL of a CaSO4lag) solution but then one of his graduate student accidentally poured in a 0.125 M solution of calcium phosphate [Ca3(PO4)2] solution. You may ignore the additional volume coming from the Cag(PO4)2 solution to make the calculation simpler. Calculate the new equilibrium constant (K) with this common ion effect. 1.20 x 10-5 3.44 x 10-5 3.84 x 10-5 2.01 x 10-5 O 1.48 x 10-5 1591....docx LAB EXP, #3 -.docx Show All
Determine the molar solubility for Agl (Ksp = 8.3 × 101"). 1 2 NEXT > Based on the given values, set up ICE table in order to determine the unknown. Agl(s) Ag*(aq) I'(aq) + Initial (M) Change (M) Equilibrium (M) 5 RESET 8.3 x 1017 -8.3 x 1017 1.66 x 1016 -1.66 х 1016 +x -X +2x -2x 8.3 x 1017 + x 8.3 x 1017 - x 8.3 x 1017 + 2x 8.3 x 1017 - 2x 1.66 x 1016 + x 1.66 x 1016 - x 1.66 x 1016 + 2x 1.66 x 1016 - 2x
could you help me understand how to do this problem and explain the steps you used please, could you also use an ICE table as well thank you.
Which of the following statements is true regarding the PbBr2 ionic equilibrium: PbBr2 (s) ⇌ Pb2+ (aq) + 2 Br- (aq) a. Adding more PbBr2(s) will cause an increase in the [Br-]. b. Adding Pb(NO3)2 will cause the [Br-] to increase. c. Adding NaBr will cause more PbBr2 to precipitate. d. None of these statements is true.