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Chapter 4: Problem 53

Exactly \(1.00 \mathrm{mL}\) of an aqueous solution of \(\mathrm{HNO}_{3}\) is diluted to \(100.0 \mathrm{mL}\). It takes \(29.78 \mathrm{mL}\) of \(0.0142 \mathrm{M}\) \(\mathrm{Ca}(\mathrm{OH})_{2}\) to convert all of the \(\mathrm{HNO}_{3}\) to \(\mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2}\) The other product of the reaction is water. Calculate the molarity of the undiluted HNO \(_{3}\) solution.

Short Answer

Expert verified
The molarity of the undiluted HNO_3 solution is 0.84552 M

Step by step solution

01

Identify the reaction equation

Determine the reaction equation between \(HNO_3\) (nitric acid), \(Ca(OH)_2\) (calcium hydroxide) and their resulting products. The balanced equation is: \(2HNO_3 + Ca(OH)_2 \rightarrow Ca(NO_3)_2 + 2H_2O\) This means that two moles of nitric acid react with one mole of calcium hydroxide to produce calcium nitrate and water.
02

Calculate the mol of \(Ca(OH)_2\)

Determine the number of mole of \(Ca(OH)_2\) used in the reaction via the formula: \[mol = Molarity \ x \ Volume\] Where the molarity is given as 0.0142 M and the volume is 29.78 mL or 0.02978L. Plugging these values into the formula gives: \[mol = \ 0.0142 M \times 0.02978 L = 0.00042276\ mol \] of \(Ca(OH)_2\)
03

Calculate the mol of \(HNO_3\)

In the balanced equation, it can be observed that 2 moles of \(HNO_3\) reacts with 1 mole of \(Ca(OH)_2\). This means that the amount of \(HNO_3\) moles is twice the amount of \(Ca(OH)_2\). So the \(HNO_3\) moles is: \[0.00042276 \ mol \times 2 = 0.00084552\ mol \] of \(HNO_3\)
04

Calculate the Molarity of undiluted \(HNO_3\)

Recall that molarity is defined as the number of moles per liter of solution. As it was stated that one ml of the acid was diluted to 100mL, the molarity of the undiluted solution \(HNO_3\), using the formula \( Molarity = \frac{mol}{Volume} \), would be: \[\frac{0.00084552}{0.001 L} = 0.84552 M\]

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Key Concepts

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

Molarity Calculation
Molarity is a fundamental concept in chemistry, referring to the concentration of a solute in a solution. It's defined as the number of moles of solute per liter of solution. This calculation is essential in titrations to determine the strength of a solution.

To find the molarity, you can use the formula:
  • \( Molarity = \frac{moles\ of\ solute}{liters\ of\ solution} \)
In this exercise, we first determine the moles of \( Ca(OH)_2 \) using its known volume and molarity. Next, using the balanced chemical equation, the moles of \( HNO_3 \) are found to be twice the moles of \( Ca(OH)_2 \). Finally, we calculate the molarity of the original \( HNO_3 \) solution by dividing its moles by the original volume in liters. This gives a direct insight into how concentrated our initial solution is before any dilution. Remember that molarity changes with dilution but the number of moles of solute remains the same.
Balanced Chemical Equations
Chemical reactions are best described using balanced chemical equations. These equations provide critical information about the proportion in which reactants combine and products are formed. For instance, in the balanced equation given:
  • \( 2HNO_3 + Ca(OH)_2 \rightarrow Ca(NO_3)_2 + 2H_2O \)
We see that two moles of nitric acid are required to fully react with one mole of calcium hydroxide.

Balancing equations ensures that the law of conservation of mass is adhered to, meaning atoms are neither created nor destroyed in a chemical reaction. To balance an equation, adjust the coefficients in front of the chemical formulas until the number of each type of atom is the same on both sides.

Importantly, understanding the stoichiometry of a balanced reaction allows for the calculation of reactants or products needed or formed in a given quantity.
Dilution of Solutions
Dilution is a common practice in chemistry when working with solutions. It involves adding solvent to a solution, thereby decreasing its concentration, while the amount of solute remains unchanged.

The relationship between the initial and final concentrations and volumes during a dilution is given by the equation:
  • \( M_1 \times V_1 = M_2 \times V_2 \)
Where \( M_1 \) and \( V_1 \) are the molarity and volume of the initial solution, and \( M_2 \) and \( V_2 \) are those of the diluted solution.

In our exercise, the volume of the \( HNO_3 \) solution is increased from 1 mL to 100 mL. Despite this dilution, the total number of moles of acid remains the same, illustrating that dilution only affects concentration, not the amount of solute present.

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Most popular questions from this chapter

When water and methanol, \(\mathrm{CH}_{3} \mathrm{OH}(\mathrm{l}),\) are mixed, the total volume of the resulting solution is not equal to the sum of the pure liquid volumes. (Refer to Exercise 99 for an explanation.) When \(72.061 \mathrm{g} \mathrm{H}_{2} \mathrm{O}\) and \(192.25 \mathrm{g}\) \(\mathrm{CH}_{3} \mathrm{OH}\) are mixed at \(25^{\circ} \mathrm{C},\) the resulting solution has a density of \(0.86070 \mathrm{g} / \mathrm{mL} .\) At \(25^{\circ} \mathrm{C},\) the densities of water and methanol are \(0.99705 \mathrm{g} / \mathrm{mL}\) and \(0.78706\) \(\mathrm{g} / \mathrm{mL},\) respectively. (a) Calculate the volumes of the pure liquid samples and the solution, and show that the pure liquid volumes are not additive. [ Hint: Although the volumes are not additive, the masses are.] (b) Calculate the molarity of \(\mathrm{CH}_{3} \mathrm{OH}\) in this solution.

Titanium tetrachloride, \(\mathrm{TiCl}_{4}\) is prepared by the reaction below. $$\begin{aligned} &3 \mathrm{TiO}_{2}(\mathrm{s})+4 \mathrm{C}(\mathrm{s})+6 \mathrm{Cl}_{2}(\mathrm{g}) \longrightarrow 3 \mathrm{TiCl}_{4}(\mathrm{g})+2 \mathrm{CO}_{2}(\mathrm{g})+2 \mathrm{CO}(\mathrm{g}) \end{aligned}$$ What is the maximum mass of \(\mathrm{TiCl}_{4}\) that can be obtained from \(35 \mathrm{g} \mathrm{TiO}_{2^{\prime}} 45 \mathrm{g} \mathrm{Cl}_{2^{\prime}}\) and \(11 \mathrm{g} \mathrm{C} ?\)

The minerals calcite, \(\mathrm{CaCO}_{3},\) magnesite, \(\mathrm{MgCO}_{3}\) and dolomite, \(\mathrm{CaCO}_{3} \cdot \mathrm{MgCO}_{3},\) decompose when strongly heated to form the corresponding metal oxide(s) and carbon dioxide gas. A 1.000 -g sample known to be one of the three minerals was strongly heated and \(0.477 \mathrm{g} \mathrm{CO}_{2}\) was obtained. Which of the three minerals was it?

A \(25.00 \mathrm{mL}\) sample of \(\mathrm{HCl}(\mathrm{aq})\) was added to a \(0.1000 \mathrm{g}\) sample of \(\mathrm{CaCO}_{3}\). All the \(\mathrm{CaCO}_{3}\) reacted, leaving some excess HCl(aq). \(\mathrm{CaCO}_{3}(\mathrm{s})+2 \mathrm{HCl}(\mathrm{aq}) \longrightarrow\) $$ \mathrm{CaCl}_{2}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l})+\mathrm{CO}_{2}(\mathrm{g}) $$ The excess HCl(aq) required 43.82 mL of 0.01185 M \(\mathrm{Ba}(\mathrm{OH})_{2}\) to complete the following reaction. What was the molarity of the original HCl(aq)? $$2 \mathrm{HCl}(\mathrm{aq})+\mathrm{Ba}(\mathrm{OH})_{2}(\mathrm{aq}) \longrightarrow \mathrm{BaCl}_{2}(\mathrm{aq})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{l})$$

Nitrogen gas, \(\mathrm{N}_{2}\), can be prepared by passing gaseous ammonia over solid copper(II) oxide, \(\mathrm{CuO}\), at high temperatures. The other products of the reaction are solid copper, \(\mathrm{Cu},\) and water vapor. In a certain experiment, a reaction mixture containing \(18.1 \mathrm{g} \mathrm{NH}_{3}\) and \(90.4 \mathrm{g}\) CuO yields \(6.63 \mathrm{g} \mathrm{N}_{2}\). Calculate the percent yield for this experiment.
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