Concept explainers
An analytical procedure for finding the chloride ion concentration in a solution involves the precipitation of silver chloride:
Silver nitrate solution is added to an unknown chloride-ion solution, yielding a precipitate of silver chloride.
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Chapter 16 Solutions
Introductory Chemistry: An Active Learning Approach
- A student is asked to identify the metal nitrate present in an aqueous solution. The cation in the solution can be either Na+, Ba2+, Ag+, or Ni2+. Results of solubility experiments are as follows: unknown + chloride ions—no precipitate unknown + carbonate ions—precipitate unknown + sulfate ions—precipitate What is the cation in the solution?arrow_forwardA soluble iodide was dissolved in water. Then an excess of silver nitrate, AgNO3, was added to precipitate all of the iodide ion as silver iodide, AgI. If 1.545 g of the soluble iodide gave 2.185 g of silver iodide, how many grams of iodine are in the sample of soluble iodide? What is the mass percentage of iodine, I, in the compound?arrow_forwardSodium chloride is used in intravenous solutions for medical applications. The NaCl concentration in such solutions must be accurately known and can be assessed by reacting the solution with an experimentally determined volume of AgNO3 solution of known concentration. The net ionic equation is Ag+(aq)+Cl(aq)AgCl(s) Suppose that a chemical technician uses 19.3 mL of 0.200-M AgNO3 to convert all the NaCl in a 25.0-mL sample of an intravenous solution to AgCl. Calculate the molarity of NaCl in the solution.arrow_forward
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- Arsenic acid, H3AsO4, is a poisonous acid that has been used in the treatment of wood to prevent insect damage. Arsenic acid has three acidic protons. Say you take a 25.00-mL sample of arsenic acid and prepare it for titration with NaOH by adding 25.00 mL of water. The complete neutralization of this solution requires the addition of 53.07 mL of 0.6441 M NaOH solution. Write the balanced chemical reaction for the titration, and calculate the molarity of the arsenic acid sample.arrow_forwardYou pour 150.0 mL of a 0.250 M lead(ll) nitrate solution into an empty 500-mL flask. What is the concentration of nitrate ions in the solution? What volume of 0.100 M sodium phosphate must be added to precipitate the lead(ll) ions from the solution?arrow_forwardYou mix 25.0 mL of 0.234-M FeCl3 solution with 42.5 mL of 0.453-M NaOH. Calculate the maximum mass, in grams, of Fe(OH)3 that will precipitate. Determine which reactant is in excess. Calculate the concentration of the excess reactant remaining in solution after the maximum mass of Fe(OH)3 has precipitated.arrow_forward
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