(a)
Interpretation: which cation present in the unknown soluble ionic compound has to be predicted.
Concept introduction: When the cations and anions switch partners, they result in the arrangement of two new ionic compounds one of which is in the solid state. The hard product is an insoluble ionic compound called a precipitate.
The rules for salts to be soluble in water are as follows
- 1. Most of the Nitrate salts are soluble
- 2. Salts containing the cation as alkali metal such as Sodium, Magnesium, Rubidium etc and Ammonium are said to be water soluble.
- 3. Salts of Bromide, Chloride and Iodide are said to be water soluble except in case of cations such as silver, Lead and Mercury.
- 4. Most of the Sulphate salts are soluble except Barium sulphate, Mercury sulphate, Lead sulphate and Calcium sulphate.
- 5. Most of the hydroxides are sparingly soluble except Sodium hydroxide and Potassium hydroxide.
- 6. Most of the Sulphide, Carbonate, Chromates and Phosphates are less soluble except those include in the rule 2.
(b)
Interpretation: which cation present in the unknown soluble ionic compound has to be predicted.
Concept introduction: When the cations and anions switch partners, they result in the arrangement of two new ionic compounds one of which is in the solid state. The hard product is an insoluble ionic compound called a precipitate.
The rules for salts to be soluble in water are as follows
- 1. Most of the Nitrate salts are soluble
- 2. Salts containing the cation as alkali metal such as Sodium, Magnesium, Rubidium etc and Ammonium are said to be water soluble.
- 3. Salts of Bromide, Chloride and Iodide are said to be water soluble except in case of cations such as silver, Lead and Mercury.
- 4. Most of the Sulphate salts are soluble except Barium sulphate, Mercury sulphate, Lead sulphate and Calcium sulphate.
- 5. Most of the hydroxides are sparingly soluble except Sodium hydroxide and Potassium hydroxide.
- 6. Most of the Sulphide, Carbonate, Chromates and Phosphates are less soluble except those include in the rule 2.
(c)
Interpretation: which cation present in the unknown soluble ionic compound has to be predicted.
Concept introduction: When the cations and anions switch partners, they result in the arrangement of two new ionic compounds one of which is in the solid state. The hard product is an insoluble ionic compound called a precipitate.
The rules for salts to be soluble in water are as follows
- 1. Most of the Nitrate salts are soluble
- 2. Salts containing the cation as alkali metal such as Sodium, Magnesium, Rubidium etc and Ammonium are said to be water soluble.
- 3. Salts of Bromide, Chloride and Iodide are said to be water soluble except in case of cations such as silver, Lead and Mercury.
- 4. Most of the Sulphate salts are soluble except Barium sulphate, Mercury sulphate, Lead sulphate and Calcium sulphate.
- 5. Most of the hydroxides are sparingly soluble except Sodium hydroxide and Potassium hydroxide.
- 6. Most of the Sulphide, Carbonate, Chromates and Phosphates are less soluble except those include in the rule 2.
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Chapter 6 Solutions
Chemistry: An Atoms First Approach
- A student added 50.0 mL of an NaOH solution to 100.0 mL of 0.400 M HCl. The solution was then treated with an excess of aqueous chromium(III) nitrate, resulting in formation of 2.06 g of precipitate. Determine the concentration of the NaOH solution.arrow_forwardChromium(III) chloride forms many compounds with ammonia. To find the formula of one of these compounds, you titrate the NH3 in the compound with standardized acid. Cr(NH3)xCl3(aq) + x HCl(aq) x NH4+(aq) + Cr3+(aq) + (x + 3) Cl(aq) Assume that 24.26 mL of 1.500 M HCl is used to titrate 1.580 g of Cr(NH3)xCl3. What is the value of x?arrow_forwardThe amount of oxygen, O2, dissolved in a water sample at 25 C can be determined by titration. The first step is to add solutions of MnSO4 and NaOH to the water to convert the dissolved oxygen to MnO2. A solution of H2SO4 and KI is then added to convert the MnO2 to Mn2+, and the iodide ion is converted to I2. The I2 is then titrated with standardized Na2S2O3. (a) Balance the equation for the reaction of Mn2+ ions with O2 in basic solution. (b) Balance the equation for the reaction of MnO2 with I in acid solution. (c) Balance the equation for the reaction of S2O32 with I2. (d) Calculate the amount of O2 in 25.0 mL of water if the titration requires 2.45 mL of 0.0112 M Na2S2O3 solution.arrow_forward
- One method for determining the purity of aspirin (C9H8O4) is to hydrolyze it with NaOH solution and then to titrate the remaining NaOH. The reaction of aspirin with NaOH is as follows: A sample of aspirin with a mass of 1.427 g was boiled in 50.00 mL of 0.500 M NaOH. After the solution was cooled, it took 31.92 mL of 0.289 M HCl to titrate the excess NaOH. Calculate the purity of the aspirin. What indicator should be used for this titration? Why?arrow_forwardWhat mass of solid aluminum hydroxide can be produced when 50.0 mL of 0.200 M Al(NO3)3 is added to 200.0 mL of 0.100 M KOH?arrow_forwardConsider an experiment in which two burets, Y and Z, are simultaneously draining into a beaker that initially contained 275.0 mL of 0.300 M HCl. Buret Y contains 0.150 M NaOH and buret Z contains 0.250 M KOH. The stoichiometric point in the titration is reached 60.65 minutes after Y and Z were started simultaneously. The total volume in the beaker at the stoichiometric point is 655 mL. Calculate the flow rates of burets Y and Z. Assume the flow rates remain constant during the experiment.arrow_forward
- Two students titrate different samples of the same solution of HCl using 0.100 M NaOH solution and phenolphthalein indicator (Figure 4.12). The first student pipets 20.0 mL of the HCl solution into a flask, adds 20 mL of distilled water and a few drops of phenolphthalein solution, and titrates until a lasting pink color appears. The second student pipets 20.0 mL of the HCl solution into a flask, adds 60 mL of distilled water and a few drops of phenolphthalein solution, and titrates to the first lasting pink color. Each student correctly calculates the molarity of an HCl solution. What will the second students result be? (a) four times less than the first students result (b) four times greater than the first students result (c) two times less than the first students result (d) two times greater than the first students result (e) the same as the first students resultarrow_forwardConsider a 1.50-g mixture of magnesium nitrate and magnesium chloride. After dissolving this mixture in water, 0.500 M silver nitrate is added dropwise until precipitate formation is complete. The mass of the white precipitate formed is 0.641 g. a. Calculate the mass percent of magnesium chloride in the mixture. b. Determine the minimum volume of silver nitrate that must have been added to ensure complete formation of the precipitate.arrow_forwardThe cations Ba2+ and Sr2+ can be precipitated as very insoluble sulfates. (a) If you add sodium sulfate to a solution containing these metal cations, each with a concentration of 0.10 M, which is precipitated first, BaSO4 or SrSO4? (b) What will be the concentration of the first ion that precipitates (Ba2+ or Sr2+) when the second, more soluble salt begins to precipitate?arrow_forward
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