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You add \(50.0 \mathrm{~mL}\) of \(0.100 M \mathrm{HCl}\) to \(50.0 \mathrm{~mL}\) of \(0.100 M\) \(\mathrm{AgNO}_{3}\). What are the final concentrations of \(\mathrm{H}_{3} \mathrm{O}^{+}\) and \(\mathrm{Cl}^{-}\) in the solution?

Short Answer

Expert verified
[\mathrm{H}_3\mathrm{O}^+] = 0.050 \text{ M} \\[\mathrm{Cl}^-] = 0 \text{ M}

Step by step solution

01

Write the initial equation and calculate moles

The reaction between hydrochloric acid \(\mathrm{HCl}\) and silver nitrate \(\mathrm{AgNO}_3\) is given by the equation: \[ \mathrm{HCl} + \mathrm{AgNO}_3 \rightarrow \mathrm{AgCl} + \mathrm{HNO}_3 \] To find the moles of each reactant, we use the formula: \[ \text{Moles} = \text{Volume in L} \times \text{Molarity} \] For \(\mathrm{HCl}\): \[ 0.050 \text{ L} \times 0.100 \text{ M} = 0.005 \text{ mol} \] For \(\mathrm{AgNO}_3\): \[ 0.050 \text{ L} \times 0.100 \text{ M} = 0.005 \text{ mol} \]
02

Determine limiting reactant and reaction progress

The reaction is in a 1:1 molar ratio, and both \(\mathrm{HCl}\) and \(\mathrm{AgNO}_3\) have the same number of moles (0.005 mol). Therefore, all reactants will be fully consumed in the reaction.
03

Calculate final concentrations in the solution

All \(\mathrm{HCl}\) and \(\mathrm{AgNO}_3\) react to form \(\mathrm{AgCl}\) and \(\mathrm{HNO}_3\). Now, the total volume of the solution is the sum of the individual volumes: \[ 50.0 \text{ mL} + 50.0 \text{ mL} = 100.0 \text{ mL} = 0.100 \text{ L} \] Since \(\mathrm{AgCl}\) precipitates out, it leaves only \(\mathrm{HNO}_3\) in solution for hydronium ions. The moles of \(\mathrm{H}_3\mathrm{O}^+\) derive from the full dissociation of the \(\mathrm{HNO}_3\), keeping the same concentration as the initial \(\mathrm{HCl}\) solution. Thus, \[ \text{Concentration of } \mathrm{H}_3\mathrm{O}^+ = \frac{0.005 \text{ mol}}{0.100 \text{ L}} = 0.050 \text{ M} \] The \(\mathrm{Cl}^-\) ions form a solid due to precipitation with \(\mathrm{Ag}^+\), resulting in their effective concentration in the aqueous solution being zero.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Limiting Reactant
In any chemical reaction, identifying the limiting reactant is crucial since it determines how much product can be formed. The limiting reactant is the substance that is fully consumed first, causing the reaction to stop. In this exercise, we see a reaction between hydrochloric acid (HCl) and silver nitrate (AgNO鈧), which are mixed in equal molar amounts.
To find the limiting reactant, we first calculate the moles of each reactant. For both HCl and AgNO鈧, we multiply their volume in liters (0.050 L) by their molarity (0.100 M), yielding 0.005 moles each.
Since the chemical equation shows that HCl and AgNO鈧 react in a 1:1 molar ratio, and since both have equal moles, neither is in excess. Therefore, both are limiting reactants in this scenario, and both are entirely used up during the reaction.
Molarity Calculation
Molarity, a fundamental concept in chemistry, measures the concentration of a solution. It is defined as the number of moles of solute per liter of solution. In this problem, we begin by finding the molarity of hydronium ions ( \(\mathrm{H}_3\mathrm{O}^+\)), formed from the full dissociation of nitric acid (HNO鈧).
We know that when HCl reacts with AgNO鈧, it forms AgCl and HNO鈧, with all HCl being converted to HNO鈧. Since the initial concentration of HCl is 0.005 moles in 50 mL, we convert the total volume to liters (100 mL = 0.100 L) for calculation of the final molarity:
  • Moles of \(\mathrm{H}_3\mathrm{O}^+\) = 0.005 moles (from complete dissociation of HNO鈧)
  • Total volume = 0.100 L
  • Molarity = \(\frac{0.005}{0.100} = 0.050 \mathrm{M}\)
This demonstrates the direct relationship between molarity, moles, and volume, reflecting how concentrated the hydronium ions are in the final solution.
Precipitation Reaction
A precipitation reaction in chemistry involves the formation of an insoluble solid, known as the precipitate. This type of reaction occurs when solutions containing soluble ions are mixed, and an insoluble compound is formed.
In the outlined exercise, the reaction between HCl and AgNO鈧 leads to the formation of silver chloride (AgCl), an insoluble compound, which precipitates out of the solution. This means that Ag鈦 and Cl鈦 ions combine to form solid AgCl.
The process can be summarized as such:
  • When the moles of Ag鈦 ions from AgNO鈧 meet the moles of Cl鈦 ions from HCl, AgCl is formed.
  • AgCl falls out of the solution as a solid; thus, the concentration of Cl鈦 ions in the remaining aqueous solution is effectively zero post-reaction.
The precipitation of AgCl is pivotal in affecting the final ion concentrations in the solution, illustrating the significance of precipitation reactions in determining solution composition.

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Most popular questions from this chapter

Suppose the \(K_{a}\) values for the hypothetical acids HX, HY, and \(\mathrm{HZ}\) are \(9.5 \times 10^{-5}, 7.6 \times 10^{-4}\), and \(1.2 \times 10^{-2}\), respectively. Assuming that "A" is a cation that produces a slightly soluble salt with all three acids, which of the following salts would have its solubility most affected (compared with its solubility in pure water) when it is dissolved in a \(0.10 M\) solution of \(\mathrm{HCl}\) ? a. \(\mathrm{AX}\) b. \(\mathrm{AY}\) c. \(\mathrm{AZ}\) d. The solubility of all the salts would be affected by the same amount. e. There is not enough information given in the problem to determine the answer.

A student dissolved a compound in water and added hydrochloric acid. No precipitate formed. Next she bubbled \(\mathrm{H}_{2} \mathrm{~S}\) into this solution, but again no precipitate formed. However, when she made the solution basic with ammonia and bubbled in \(\mathrm{H}_{2} \mathrm{~S}\), a precipitate formed. Which of the following are possible as the cation in the compound? a. \(\mathrm{Ag}^{+}\) b. \(\mathrm{Ca}^{2+}\) C. \(\mathrm{Mn}^{2+}\) d. \(\mathrm{Cd}^{2+}\)

The solubility of magnesium oxalate, \(\mathrm{MgC}_{2} \mathrm{O}_{4}\), in water is \(0.0093 \mathrm{~mol} / \mathrm{L} .\) Calculate \(K_{s p}\)

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A 45-mL sample of \(0.015 M\) calcium chloride, \(\mathrm{CaCl}_{2}\), is added to \(55 \mathrm{~mL}\) of \(0.010 M\) sodium sulfate, \(\mathrm{Na}_{2} \mathrm{SO}_{4}\). Is a precipitate expected? Explain.

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