What is the molar solubility of barium oxalate(BaC2O4) in a solution buffered to a pH of 6.65? The Ksp for BaC2O4 is 2.3 x 10^-8 and the Ka's for H2C2O4 are Ka1 = 5.9 X 10^-2 and Ka2 = 6.4 x 10^-5. (Write over all equilibrium equation and calculate K for doing equilibrium.)

A chemistry graduate student is given 100. mL of a 1.30M benzoic acid (HC,H,CO,) solution. Benzoic acid is a weak acid with K,= 6.3 × 10 °. What mass of NaC,H,CO, should the student dissolve in the HC,H,CO, solution to turn it into a buffer with pH = 4.01? You may assume that the volume of the solution doesn't change when the NaCH,CO, is dissolved in it. Be sure your answer has a unit symbol, and round it to 2 significant digits. x10

What is the molar solubility of calcium sulfate, CaSO4, at 25°C? The solubility product constant for calcium sulfate is 2.4  10–5at 25°C.A) 2.4  10–5 MB) 3.5  10–2 MC) 1.2  10–5 MD) 1.8  10–2 ME) 4.9  10–3 M

In order to determine the chloride concentration in a solution, a Mohr titration is perform. This involves adding a small amount of chromate to the solution and then slowly adding silver nitrate. Silver chloride is pale yellow solid while Silver chromate is a dark red. The appearance of the red silver chromate is the endpoint of the titration. Answer the questions below about this process. (Ksp of AgCl= 1.7x10^-10 and Ksp of Ag2CrO4= 1.12x10^-12) A 100 mL sample of water from a potentially contaminated stream is collected. A 0.0010 g of sodium chromate is dissolved in the sample. 1) if the solution had been 0.015 M in NaCl, what percent of the chloride is still in solution?  Given the answer to the question above, it should be clear that "all" of the chloride is removed before the silver chromate appears. This is why we can use it as an endpoint. 2) if it requires 21.4 mL of a 0.1 M solution of silver nitrate to reach the equivalence point with 100 mL of solution, what was the original…

What is the molar solubility of Fe(OH)3 in a solution buffered at pH = 4.29? See the Solubility Product Constant Table. 4.0 8.1e-9 What is the molar solubility at pH = 9.00? 4.0 6.00-23 X M Substance Ksp at 25 °C Aluminum Al(OH)3 1.9 x 10-33 Barium Ba(OH)2 BaCO3 BaSO4 XM 5.0 x 10-3 8.1 x 10-9 1.1 x 10-10 Ba3(PO4)2 3.4 x 10-23 Cadmium Cd(OH)2 CdCO3 CdS Calcium 2.5 x 10-14 5.2 x 10-12 8.0 x 10-27 Ca(OH)2 5.5 x 10-6 CaCO3 4.8 x 10-⁹ Ca3(PO4)2 1.0 x 10-26 CaF2 3.9 x 10-11 Chromium CO(OH)2 COS Cr(OH)3 6.3 x 10-31 Cobalt 1.6 × 10-15 4.0 x 10-21 Copper Cu(OH)2 Cu₂S CUCO3 CuS Iron Fe(OH)2 FeS Fe(OH)3 FePO4 Lead Pb(OH)2 PbF2 PbCl₂ PbBr₂ Pbl2 PbCrO4 PbSO4 PbS 2.2 x 10-20 1.6 x 10-48 1.4 x 10-10 6.3 x 10-36 Mg(OH)2 8.0 x 10-16 6.3 x 10-18 2.5 x 10-39 9.9 x 10-29 2.8 x 10-16 3.7 x 10-8 1.7 x 10-5 6.3 x 10-6 6.5 x 10-⁹ 1.8 x 10-14 1.7 x 10-8 8.4 x 10-28 Magnesium 1.8 × 10-11 Manganese Mn(OH)2 MnCO3 MnS Nickel Ni(OH)2 Silver AgCl AgBr Agl AgCN Ag₂S Ag₂ CrO4 Ag3PO4 Tin Sn(OH)2 SnS Zinc ZnCO3 Zn(OH)2…

If 200 mL of 1.0 x 10^-8 M AgNO_3 is mixed with 200 mL of 1.0 x 10^-8 M Nal, what will occur? For Agl, Ksp = 8.5 x 10-17, Sodium iodide will precipitate. Sodium nitrate will precipitate. No precipitate will form. Silver iodide will precipitate Silver nitrate will precipitate.

The solubility of Sr(OH)2 is 3.2 x 104 g/L a) Write a balanced equation for the solubility equilibrium b) Write the expression for the solubility product constant Ksp and calculate its value c) Calculate the pH of a saturated solution of Sr(OH) 2 d) 25 ml of 3 x 10-4 SOLUTION of SrS04 is added to 35 ml of 3.21 x 10-6 molar NaH solution. Does a precipitate form? Explain and show calculations to support your answer

The solubility of Ca(OH)2 is 1.3 x 10-5 g/L a) Write a balanced equation for the solubility equilibrium b) Write the expression for the solubility product constant Ksp and calculate its value c) Calculate the pH of a saturated solution of Ca(OH)2 d) 50 ml of 1.7 x 10-4 SOLUTION of CaS04 is added to 150 ml of 6.8 x 10-6 molar NaH solution. Does a precipitate form? Explain and show calculations to support your answer

A buffer is made with 0.50 M weak acid and 0.50 M sodium salt of the conjugate base. The pKa of the weak acid is 9.30 After the addition of a small amount of strong concentrated acid, the new equilibrium concentrations are 0.6 M weak acid and 0.4 M base. Calculate the new pH. Enter result with 2 decimal places, digits only.

Potassium hydroxide is used to precipitate each of the cations from their respective solution. Determine the minimum concentration of KOHKOH required for precipitation to begin in each case. 1. 0.010 MM CaCl2CaCl2 (Ksp=4.68×10−6Ksp=4.68×10−6 for Ca(OH)2Ca(OH)2) 2. 0.0020 MM Fe(NO3)2Fe(NO3)2 (Ksp=4.87×10−17Ksp=4.87×10−17 for Fe(OH)2Fe(OH)2) 3. 0.0016 MM MgBr2MgBr2 (Ksp=2.06×10−13Ksp=2.06×10−13 for Mg(OH)2Mg(OH)2)