Dibasic calcium phosphate (CaHP0_4) has limited solubility in water with solubility product of K_spCaHPO_4 = 7-23...
Calcium hydroxide and lead chloride both have limited solubility in neutral water. When acid is added, the solubility of calcium hydroxide goes way up. but that of lead chloride doesn't change. Which statement explains the acid impact? Ca(OH)_2(s) Ca^2+ (aq) + 2 HO^(aq) K = 4.7 x 10 - 6 PbCl_2(s) Pb^2+ (aq) + 2 Cl-(aq) K= 1.6 times 10-5 HO-(aq) + H+(aq) - H2O(1) a Acid reacts with the calcium ions that are produced. b. Acid increases the solubility...
Write a solubility product constant expression, K_sp, for an equilibrium in a saturated aqueous solution of the slightly soluble salts Magnesium fluoride, MgF_2 Zinc phosphate, Zn_3(PO_4)_2 Chromium(II) hydroxide, Cr(OH)_3 Strontinum sulfate, SrSO_4. Calculate the solubility in g/L for calcium carbonate, CaCO_3, in pure water given the K_sp = 2.8 times 10^-9. Calculate the solubility in g/L for CaCO_3, in 0.050 M Na_2CO_3.
2. The equilibrium constant for the following reaction is called the "solubility product" of calcium fluoride: CaF2(s) - Cal(aq) + 2F-(ay) K = Kp = 3.2 x 10-11 (a) Write an expression for the equilibrium constant of this reaction in terms of concentrations. Why do you suppose we call this a solubility product instead of a solubility quotient or ratio? (b) Calculate the equilibrium concentrations of Ca2+ and F if excess solid CaF is placed in water. (c) in which...
A student dissolves calcium carbonate solids, CaCO_3(s) in distilled water. K_sp = 5.0 times 10^-9 a) Calculate the molar solubility of calcium carbonate in distilled water. Express your answer in gram per litre b) How would the answers in a) change if the student dissolved CaCO_3 in 0.01 mol/L solution of Na_2CO_3(aq) c) The student mixes 0.0140 g of CaO solids with 0.0265 g of Na_2CO_3 solids in 1.0 L of water solids are soluble in water). Predict whether precipitate...
help me out, please! answer all the multiple choice. (15) A phosphate buffer solution is prepared by mixing 100. mL. of 0.300 M KH PO, and 150.ml 0.500 M K HPO (a) Calculate the molar concentrations of H:PO, and that of HPO,2 in the buffer solution. (b) What is the pH of the buffer solution? (H,PO, has K,-6.2 x 10 (c) Write a net ionic equation for the bufering reaction against a strong acid, Hjo (d) Calculate the new concentration...
4.30. Using solubility rules, predict the solubility in water of the following ionic compounds. a. AI(OH) b. CaN C. NH4CI d. KOH 4.32. Using solubility rules, decide whether the following ionic solids are soluble or insoluble in water. If they are soluble, write the chemical equation for dissolving in water and indicate what ions you would expect to be present in solution. (NE SO b. BaCO c. Pb(NOs)2 d. Ca(OH) 4.34. Write net ionic equations for the following molecular equations....
15) Determine the molar solubility of FeCl, in pure water. K A) 0.228 M B ) 0.0301 M C) 0.311 M for FeCl, -2.45 x 10 D ) 0.174 M 15 E) 2.45 * 10M 16) 16) Which of the following statements is true? A) Kris related to the formation of complexes B) K is related to acid dissociation C) K, is related to the solubility of a solid ionic species D) Khis related to base dissociation E) K is...
ALT CHEM 125 Name Determining the Ksp of Calcium Hydroxide Calcium hydroxide is an ionie solid that is sparingly soluble in water. A saturated, aqueous, solution of Ca(OH), is represented in equation form as shown below. Ca(OH)2 (5) --- Ca' (aq) + 2OH(aq) The solubility product expression describes, in mathematical terms, the equilibrium that is established between the solid substance and its dissolved ions in an aqueous system. The equilibrium expression for calcium hydroxide is shown below. Kp - [Ca][OHT...
Part A and Part B. Also question #3-post lab. (its circled) A. Molar Solubility and Solubility Product of Calcium Hydroxide Trial I Trial 3 Trial 2 25.0 1. Volume of saturated Ca(OH), solution (mL) 2. Concentration of standardized HCl solution (molU/L) 3. Buret reading, initial (mL 4. Buret reading, final (mL) 5. Volume of HCI added (mL) 6. Moles of HCI added (mol) 7. Moles of OH" in saturated solution (mol) 8. (OH1, equilibrium (mol/L) 9. (Ca2 ], equilibrium (mol/L)...
Experiment 8 Double Replacement Reactions Background: Some reactions have the net effect of causing the cation of each reactant to trade places, forming a compound with the other anion. These reactions are known as double replacement reactions. In the example below (unbalanced equation), the barium and sodium cations switch places so that barium forms a product with sulfate while sodium forms a product with chloride. Note that the formula of each product is determined by the charges of the ions,...