50.00 ml. of a 0.10 M NaCl (aq) is added to 75.00 mL of a 0.25 M AgNO_3 solution

Question

50.00 ml. of a 0.10 M NaCl (aq) is added to 75.00 mL of a 0.25 M AgNO_3 solution a. Write the balanced chemical equation for this reaction. b. Write the total ionic and net ionic equations. c. What are the spectator ions? d. What precipitate is formed? e. What is the concentration of the sodium ion in the resulting solution. f. What is the concentration of the nitrate ion in the resulting solution? g. What is the approximate concentration of the chloride ion in the resulting solution? h. what is the approximate concentration of the silver ion in the resulting solution?

Explanation / Answer

5a. NaCl(aq). + AgNO3(aq). ------------->. AgCl(s). + NaNO3(aq)

5b. Net ionic equation is

Cl-(aq) +. At+(aq) -------------------> AgCl.(s)

The total ionic equation is

Na+(aq). + Cl- (aq) + Ag +(aq) + NO3-( aq) .--------->. Na+(aq) + NO3-(aq) + AgCl (s)

5c. The spectator ion is an ion exists in the same form on both the reactant and product sides of the chemical reaction. For example in the current reaction the spectator ions are Na+, Cl-, Ag+, and NO3-.

5d. The precipitate formed is AgCl, silver monochloride.

5e. Conc of sodium ions in the resulting solution is 0.005 moles.

5f. Conc of nitrate ions in the resulting solution is 0.01875 moles.

5g. The approximate concentration of chloride ions in the resultant solution is 0.01375 moles.

5h. The approximate concentration of silver ions in the resultant solution is 0.01375 moles.

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