Concept explainers
(a)
Interpretation: The chloride salt of lead is converted to its chromate salt to test its presence, the reason has to be confirmed.
Concept introduction: Metal ions present in a solution can be identified by a set procedure known as qualitative analysis. Different metal ions show different reactions with different reagent and thus either through the color of the precipitate formed or by smell the gas evolved the metal ion present can be identified.
Some metal ions when present in an aqueous solution containing anions or neutral species called Lewis base or ligands having a tendency to donate electron pairs to metal ions then complex ion formation will take place. Complex ions are stable and thus formation of these increase the solubility of the salt containing the metal ions same as in complex ions.
Example of metal ions that form complex ions includes
Example of Lewis bases includes,
(b)
Interpretation: The presence of
Concept introduction: Metal ions present in a solution can be identified by a set procedure known as qualitative analysis. Different metal ions show different reactions with different reagent and thus either through the color of the precipitate formed or by smell the gas evolved the metal ion present can be identified.
Some metal ions when present in an aqueous solution containing anions or neutral species called Lewis base or ligands having a tendency to donate electron pairs to metal ions then complex ion formation will take place. Complex ions are stable and thus formation of these increase the solubility of the salt containing the metal ions same as in complex ions.
Example of metal ions that form complex ions includes
Example of Lewis bases includes,
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Chemistry & Chemical Reactivity, Hybrid Edition (with OWLv2 24-Months Printed Access Card)
- The following concentrations are found in mixtures of ions in equilibrium with slightly soluble solids. From the concentrations given, calculate Ksp for each of the slightly soluble solids indicated:(a) AgBr: [Ag+] = 5.7 × 10–7 M, [Br–] = 5.7 × 10–7 M(b) CaCO3: [Ca2+] = 5.3 × 10–3 M, [CO32−] = 9.0 × 10–7 M(c) PbF2: [Pb2+] = 2.1 × 10–3 M, [F–] = 4.2 × 10–3 M(d) Ag2CrO4: [Ag+] = 5.3 × 10–5 M, 3.2 × 10–3 M(e) InF3: [In3+] = 2.3 × 10–3 M, [F–] = 7.0 × 10–3 Marrow_forward21). The solubility product for silver phosphate is Ksp = 1.8 x 10–18. What is the molar solubility of Ag3PO4? (a) Show the reaction of dissociation of Ag3PO4. (b) Show the Ksp expression for Ag3PO4. (c) Calculate the molar solubility s for this compound.arrow_forwardCalculate the Ksp of the following compounds, given their molar solubilities. (a) CuS, 7.33 × 10-19 M (b) Cu(ОН)2, 1.76 х 10-7 М (c) Mg3 (PO4)2, 6.20 × 10–6 Marrow_forward
- find present, absent and in doubt. And the reasoning of the following problem (photo)arrow_forward(ii) Consider the titration of 25.00 mL of 0.02000 M CaSO4 with 0.01000 M EDTA at pH 10.00. Write the chemical equation for this titration and calculate the conditional formation constant for this reaction. (iii) From (ii) calculate the concentration of Ca2+ and pCa2+ at the volume of 20.0 mL of EDTA added.arrow_forwardAs part of a soil analysis on a plot of land, a scientist wants to determine the ammonium content using gravimetric analysis with sodium tetraphenylborate, Na+B(C6H5)4−. Unfortunately, the amount of potassium, which also precipitates with sodium tetraphenylborate, is non‑negligible and must be accounted for in the analysis. Assume that all potassium in the soil is present as K2CO3 and all ammonium is present as NH4Cl. A 5.095 g soil sample was dissolved to give 0.500 L of solution. A 150.0 mL aliquot was acidified and excess sodium tetraphenylborate was added to precipitate both K+ and NH4+ ions completely. B(C6H5)4-+K+⟶KB(C6H5)4(s) B(C6H5)4-+NH4+⟶NH4B(C6H5)4(s) The resulting precipitate amounted to 0.269 g. A new 300.0 mL aliquot of the original solution was made alkaline and heated to remove all of the NH4+ as NH3. The resulting solution was then acidified, and excess sodium tetraphenylborate was added to give 0.129 g of precipitate. Find the mass percentages of NH4Cl and…arrow_forward
- Calculate The solubility of Ba(IO3)2 in a solution prepared by mixing 200 mL of 0.00100 M Ba(NO3)2 with 100 mL of 0.0100 M NaIO3. The Ksp for barium iodate is 1.57 x 10-9(a) Write the solubility equilibria expression for Ba(IO3)2(b) Write the Ksp expression for the reaction in (a)(c) Determine the new concentration of Barium and Iodate ions(d) Determine the limiting and excess reactant(e) Calculate the molar solubility of Barium Iodate(f) What does a small value of Ksp tell us?arrow_forwardA student mixed equal volumes of 0.2 M solutions of sulfuric acid and calcium chloride together. (a) What precipitate forms? (b) Write an equation for the equilibrium present and the Ksp expression. (c) In the resulting solution, does [SO42-] = [Ca2+]? Explain.arrow_forward. A solution was prepared by mixing 4.00mL of 2.00 x 10-3 M Fe(NO3)3 and 3.00mL of 5.00 x 10-3 M NaSCN and diluting the mixture with water to a total of 10.00mL. Use your average value of Kc to calculate the equilibrium concentration of FeSCN2+ in the mixture. [Hint: Use as many significant figures as you legitimately can in your calculations] my average you can find on the picture below PLS HELP ASAP!!arrow_forward
- A buffer solution with a pH of 4,9, consists of 0,25 mol∙dm–3 acetic acid (HC2H3O2; Ka= 1,8 x 10–5) and 0,35 mol∙dm–3 potassium acetate (KC2H3O2). A small quantity (0,050 mol) of solid KOH is added to 1 dm3 of the buffer. This quantity is so small that it does not affect the volume of the buffer. Calculate the new pH of the buffer solution after the KOH has been added.arrow_forwardPlease solve thisarrow_forward1) Listen The Ksp value for calcium sulfate [CaSO4] is 2.40 X 10-5 and a professor made 1825 mL of a CaSO4l2g) 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. O 1.20 x 10-5 3.44 x 10-5 3.84 x 10-5 2.01 x 10-5 1.48 x 10-5 91..docx LAB EXP. #3 -.docx Show Allarrow_forward
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