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If \(27.5 \mathrm{mL}\) of \(3.5 \times 10^{-2} \mathrm{N} \mathrm{Ca}(\mathrm{OH})_{2}\) solution is needed to neutralize \(10.0 \mathrm{mL}\) of nitric acid solution of unknown concentration, what is the normality of the nitric acid?

Short Answer

Expert verified
The normality of the nitric acid solution can be calculated using the formula N₂ = \(\frac{2 \times (3.5 \times 10^{-2}) \times 0.0275}{0.01}\). After performing the calculation, you will find the normality of the nitric acid solution.

Step by step solution

01

Identify the balanced equation

First, let's write the balanced equation for the reaction between calcium hydroxide and nitric acid. In this case, one molecule of calcium hydroxide reacts with two molecules of nitric acid to produce one molecule of calcium nitrate and two molecules of water: Ca(OH)₂ (aq) + 2HNO₃ (aq) → Ca(NO₃)₂ (aq) + 2H₂O (l)
02

Calculate the amount of Ca(OH)₂ in equivalents

Next, we need to calculate the equivalents of calcium hydroxide in the solution, which can be expressed using the normality and the volume of the solution. The formula to find the equivalents (Eq) is: Eqs = N × V where N is the normality and V is the volume in liters. Given the normality of calcium hydroxide (N₁) as \(3.5 \times 10^{-2}\) N and the volume (V₁) as 27.5 mL, we can calculate the equivalents of calcium hydroxide as follows: Eqs₁ = N₁ × V₁ \(Eqs_1 = (3.5 \times 10^{-2}) \times \frac{27.5}{1000}\) Eqs₁ = \((3.5 \times 10^{-2}) \times 0.0275\)
03

Calculate the amount of HNO₃ in equivalents

According to the balanced equation, one molecule of Ca(OH)₂ is equivalent to two molecules of HNO₃, so we can calculate the equivalents of HNO₃ by multiplying the equivalents of Ca(OH)₂ by the ratio between the reactants: Eqs₂ = 2 × Eqs₁ Eqs₂ = \(2 × (3.5 \times 10^{-2}) \times 0.0275\)
04

Determine the normality of HNO₃

Now we can find the normality of the nitric acid solution by dividing the equivalents of HNO₃ (Eqs₂) by the volume of the nitric acid solution (V₂), which is 10 mL: N₂ = Eqs₂ / V₂ N₂ = \(\frac{2 \times (3.5 \times 10^{-2}) \times 0.0275}{\frac{10}{1000}}\) N₂ = \(\frac{2 \times (3.5 \times 10^{-2}) \times 0.0275}{0.01}\) Once you perform the calculation, you will find the normality of the nitric acid.solution.

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Key Concepts

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

Stoichiometry in Chemistry
Stoichiometry is a branch of chemistry that involves calculations based on the relationships of reactants and products in chemical reactions. It is essentially the 'recipe' for a chemical reaction, indicating how much of each substance is needed to react with another and what quantities of products will be formed. In our exercise, we applied stoichiometry to find out the amount of nitric acid that reacts with a given amount of calcium hydroxide.

Understanding stoichiometry is crucial when predicting the outcomes of chemical reactions. Through stoichiometric calculations, chemists can determine the optimal amounts of reactants to use in order to yield the desired quantity of products, which minimizes waste and cost. It's also the basis for titrations in analytical chemistry, where stoichiometry helps in finding out the concentration of an unknown solution.
Chemical Reactions
Chemical reactions involve the transformation of substances through the breaking and forming of chemical bonds. These processes are described by chemical equations that show the reactants converting into products. The balanced chemical equation for the reaction between calcium hydroxide and nitric acid is an excellent example, showcasing how two molecules of nitric acid react with one molecule of calcium hydroxide to produce calcium nitrate and water.

To grasp these reactions, students must understand the Law of Conservation of Mass, which states that matter is neither created nor destroyed in a chemical reaction. This principle ensures that the mass and number of atoms are equal on both sides of the equation and is a fundamental aspect of balancing chemical reactions.
Acid-Base Titration
Acid-base titration is an experimental technique used to determine the concentration of an unknown acid or base by neutralizing it with a solution of known concentration, which we refer to as a standard solution. The point at which neutralization occurs is known as the equivalence point and can be indicated by a sudden change in pH, often identified with the help of an indicator.

In the context of our textbook problem, the acid-base titration technique is used to find the unknown concentration of nitric acid, by reacting it with a known volume and normality of calcium hydroxide solution. The equivalence point in a titration of a strong acid with a strong base generally occurs at pH 7. However, the specific pH at the equivalence point can vary when dealing with weak acids or bases.
Equivalent Concept
The equivalent concept in chemistry is a way of expressing the reactive capacity of a compound in a chemical reaction. It is especially useful in reactions involving ionic compounds, such as acids and bases. The normality (N) of a solution, which is the number of equivalents per liter of solution, is a component of this concept and is applied when discussing the strength of an acid or base solution.

The calculation of normality ties in with acid-base titration, as it allows us to equate the number of acid equivalents to the number of base equivalents at the equivalence point, regardless of their actual concentration in moles. This is what made it possible to solve our textbook problem for the unknown normality of the nitric acid solution, by equalizing the number of equivalents of the acid and the base.

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Most popular questions from this chapter

A certain grade of steel is made by dissolving \(5.0 \mathrm{g}\) of carbon and \(1.5 \mathrm{g}\) of nickel per \(100 . \mathrm{g}\) of molten iron. What is the mass percent of each component in the finished steel?

Many laboratories keep bottles of \(3.0 \mathrm{M}\) solutions of the common acids on hand. Given the following molarities of the concentrated acids, determine how many milliliters of each concentrated acid would be required to prepare \(225 \mathrm{mL}\) of a \(3.0 \mathrm{M}\) solution of the acid. $$ \text {Acid} \quad \quad \text {Molarity of Concentrated Reagent} $$ $$ \text {\(\mathrm{HCl}\)} \quad \quad \text {\(12.1 M\)} $$ $$ \text {\(\mathrm{HNO}_{3}\) } \quad \quad \text {\(15.9 M\)} $$ $$ \text {\(\mathrm{H}_{2} \mathrm{SO}_{4}\)} \quad \quad \text {\(18.0 M\)} $$ $$ \text {\(\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\)} \quad \quad \text {\(17.5 M\)} $$ $$ \text {\(\mathrm{H}_{3} \mathrm{PO}_{4}\) } \quad \quad \text {\(14.9 M\)} $$

Aluminum ion may be precipitated from aqueous solution by addition of hydroxide ion, forming \(\mathrm{Al}(\mathrm{OH})_{3} .\) A large excess of hydroxide ion must not be added, however, because the precipitate of \(\mathrm{Al}(\mathrm{OH})_{3} \quad\) will redissolve to form a soluble aluminum/hydroxide complex. How many grams of solid NaOH should be added to \(10.0 \mathrm{mL}\) of \(0.250 \mathrm{M}\) \(\mathrm{AlCl}_{3}\) to just precipitate all the aluminum?

Explain why the equivalent weight of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) is half the molar mass of this substance. How many hydrogen ions does each \(\mathrm{H}_{2} \mathrm{SO}_{4}\) molecule produce when reacting with an excess of \(\mathrm{OH}^{-}\) ions?

For each of the following solutions, the number of moles of solute is given, followed by the total volume of solution prepared. Calculate the molarity. a. 0.10 mol of \(\mathrm{CaCl}_{2} ; 25 \mathrm{mL}\) b. 2.5 mol of \(\mathrm{KBr} ; 2.5 \mathrm{L}\) c. 0.55 mol of \(\mathrm{NaNO}_{3} ; 755 \mathrm{mL}\) d. 4.5 mol of \(\mathrm{Na}_{2} \mathrm{SO}_{4} ; 1.25 \mathrm{L}\)

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