Analysis of dissolved sulfate concentration is based on the precipitation reaction of bariumsulfate, creating Baso4 5) (named barite) and its settling. The following water quality is given:TDS = 650 mg/I, (SO) = 40 mg/l as S.Solubility constant of BaSO:: Kip (Baso.) = 1.1 1010Determine how many mg of BaCl, should be added to 200 ml of a solution at the givenquality for completely precipitating the sulfate. That is, calculate the dose required for 99.9%removal, while assuming saturation with respect to barite is immediately reached.

Given data: Concentration of sulfate ions = 40 mg/L Ksp of BaSO4 = 1.1 x 10-10

Analysis of dissolved sulfate concentration is based on the precipitation reaction of barium sulfate, creating Baso.s (named barite) and its settling. The following water quality is given: TDS = 650 mg/l, [SO.2) = 40 mg/l as S. Solubility constant of BaSO: K (Baso.) = 1.1 10 10 Determine how many mg of BaCl, should be added to 200 ml of a solution at the given quality for completely precipitating the sulfate. That is, calculate the dose required for 99.9 % removal, while assuming saturation with respect to barite is immediately reached.

Analysis of dissolved sulfate concentration is based on the precipitation reaction of barium sulfate, creating Baso.s (named barite) and its settling. The following water quality is given: TDS = 650 mg/l, [SO.2) = 40 mg/l as S. Solubility constant of BaSO: K (Baso.) = 1.1 10 10 Determine how many mg of BaCl, should be added to 200 ml of a solution at the given quality for completely precipitating the sulfate. That is, calculate the dose required for 99.9 % removal, while assuming saturation with respect to barite is immediately reached.

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

The corect solubility product expression for A(SO,), is: A) K = (x) (2x) D) K, = (2x) (x) B)K, = (4x)(2x)? E) K, = (x) (2x) C) K, = (x)² (2x) F) K, = 2x %3D %3D O A O E O F B O D
Ksp for calcium phosphate = 2*10^(-29)
Copper(I) ions in aqueous solution react with NH, (aq) according to Cu*(aq) +2 NH,(aq) → Cu(NH,) (aq) Kf = 6.3 × 1010 Calculate the solubility (in g-L-!) of CuBr(s) (Ksp = 6.3 × 10-9) in 0.76 M NH, (aq).
10. For the following solid Ca3(PO4)2|| a. From the balanced equation for the dissolution reaction b. From the expression of the solubility product (Ksp)
128 Identity of Oxalate Salt Solubility: •) macias CF C204" Solubility Product Constant 240 Sr C₂04-H₂0 0.00433g Sr C₂O4 - H₂0 toog H₂O Kp from Solubility (Show your calculations below) Calculate the molarity of KMnO4 solution needed to titrate 50.00 mL of a saturated solution of this oxalate salt. It should take 7.50 mL of KMnO4 solution to titrate the saturated oxalate solution. m - 000904 ROG GA oval
Provide full complete solution do not round off numbers and show proper labelling and cancellations.
One mixes aqueous NaCl with aqueous AgNO3. NaCl. AgNO3and NaNO3all have appreciable solubility in water. AgCl has a solubility product constant Ksp of 1.6 x 10-10. If one mixes the NaCl solution with the AgNO3solution, one would suspect: a) no precipitate will form b) AgCl precipitate will form c) NaNO3 precipitate will form d) both AgCl and NaNO3 precipitates will form
Copper(I) ions in aqueous solution react with NH, (aq) according to Cut (aq) + 2 NH, (аq) — Си(NH, ); (aq) 3 6.3 х 1010 - Calculate the solubility (in g-L-!) of CuBr(s) (Ksp = 6.3 × 10-9) in 0.72 M NH, (aq). solubility of CuBr(s): g/L
O KINETICS AND EQUILIBRIUM Calculating the solubility of an ionic compound when a commo... Calcula alo solubility at 25 °C of CuBr in pure water and in a 0.0070M COBT, solution. Data Round both of your answers to 2 significant digits. Formula Ksp BaCro4 1.17x10-10 BaSO4 1.08x10-10 solubility in pure water: CaCO3 3.36x10-9 CaF2 3.45x10-11 solubility in 0.0070 M CoBr, solution: Co(OH)2 5.92x10-15 CuBr 6.27x10-9 CUCO3 1.4x10-10 Fe(OH)2 4.87x10-17 PBCO3 7.40x10-14 PbCr04 2.8x10-13 PBF2 3.3x10-8 Mg(OH)2 5.61x10-12 Ni(OH)2 5.48x10-16 AgBrO3 5.38x10-5 Ag2CO3 8.46x10-12 AgCl 1.77x10-10 Ag2Cr04 1.12x10-12 SrCO3 5.60x10-10 ZnC03 1.46x10-10 Zn(OH)2 3.0x10-17 AgBr 5.35x10-13 1.77x10-10 2021 McGraw-Hill Education. All Rights Reserved Check AuCl Explanation Tems of Use Pv 18
A solution is prepared by adding 0.10 mole of Ni(NH, )eCl, to 0.50 L of 3.4 M NH, Calculate Ni (NH,)," and Ni in this solution. Ksa for Ni (NH,)," 5.5 x 10, That is, Ni (NH,),"] 5.5 x 10" for the overall reaction 2+ Ni* (ag) + 6NH, (aq) Ni (NH, )," (ag) [NI (NH,),"]-C M. Submit Answer Try Another Versian 1 item attempt remaining
3.02 g of sodium sulfate was mixed into a solution containing 48.5 mL of 0.185 mol/L aqueous silver nitrate: Calculate the maximum mass (in g) of precipitate that can form. Note: Balanced chemical eq. with state of each compound required. 2. What is the molar concentration of sodium ions in the resulting solution?
5:23 O13%DE elearning.yu.edu.jo/m O A student was given 2 beakers labeled with numbers 1 and 2, each beaker contains a clear solution. Additionally, he was given two reagent bottles labeled with A and B. Now, he added reagent A to both beakers and he noticed the formation of a precipitate in beaker 2 but not in beaker 1. However, the addition of reagent B to both beakers resulted in the formation of a precipitate in beaker 1 but not in beaker 2. Based on the information, which of the following is correct? a. compound A is the limiting reactant in beaker 1. b. compound B is the limiting reactant in beaker 2. c. compound A is the limiting reactant in beaker 2. d. compound B will be the excess reactant in beaker 1. Question 2