A 15.00-mL. sample of sodium hydroxide was titrated to the stoichiometric point with \(17.40 \mathrm{~mL}\). of \(0.234 \mathrm{M} \mathrm{HCl}(\mathrm{aq})\). (a) What is the initial molarity of \(\mathrm{NaOH}\) in the solution? (b) Calculate the mass of \(\mathrm{NaOH}\) in the solution.

Short Answer

Expert verified
The initial molarity of NaOH is 0.258M, and the mass of NaOH in the solution is 0.615g.

Step by step solution

01

Calculate moles of HCl used

Use the volume and molarity of HCl to find the moles of HCl that were used in the titration. The molarity (M) is moles per liter (mol/L), so multiply the volume by the molarity to find the moles: moles of HCl = volume of HCl (in liters) × molarity of HCl.
02

Write the balanced chemical equation

Write the balanced chemical equation for the reaction between HCl and NaOH to understand the mole ratio. The equation is: \( \text{NaOH} + \text{HCl} \rightarrow \text{NaCl} + \text{H}_2\text{O} \). The ratio of NaOH to HCl is 1:1.
03

Calculate moles of NaOH

Since the mole ratio of NaOH to HCl is 1:1, the moles of NaOH will be equal to the moles of HCl.
04

Calculate initial molarity of NaOH

The molarity (M) is moles per liter. Use the volume of NaOH in liters and the moles of NaOH calculated in Step 3 to find the molarity: molarity of NaOH = moles of NaOH / volume of NaOH (in liters).
05

Calculate the mass of NaOH

To find the mass of NaOH, use its molar mass and the moles of NaOH calculated in Step 3. The mass of NaOH (in grams) = moles of NaOH × molar mass of NaOH.

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Key Concepts

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

Stoichiometry
When it comes to understanding chemical reactions, stoichiometry is a fundamental concept that governs the quantitative relationships between reactants and products. It is essentially the 'recipe' for a chemical reaction, dictating how much of each substance is needed to react completely without any excess. Imagine baking a cake, where you need precise amounts of flour, sugar, and eggs. In the same way, stoichiometry tells you how much of one reactant is required to react with a given amount of another.

Understanding stoichiometry involves dealing with the mole concept, because reactions happen on an atomic or molecular scale. One mole is Avogadro's number of particles, which is around 6.022 x 10^23 particles - a huge number to account for the vast number of atoms or molecules involved. For example, the balanced chemical equation provided in the exercise shows a 1:1 stoichiometric ratio, meaning one mole of sodium hydroxide reacts with one mole of hydrochloric acid to produce one mole of sodium chloride and one mole of water.

To perform stoichiometric calculations, one must first interpret the balanced chemical equation, then convert quantities of one substance (usually in moles) to quantities of another substance, using the mole ratio provided by the equation.
Molarity
Molarity is one of the main ways to express the concentration of a solution. It's defined as the number of moles of solute (the dissolved substance) per liter of solution. The molarity formula is given by the expression: \( M = \frac{moles~of~solute}{liters~of~solution} \).

For instance, in the original exercise, the molarity of hydrochloric acid (HCl) is provided, allowing us to calculate the moles of HCl used in the titration. This is crucial because titration is a technique used to determine the concentration of an unknown solution by reacting it with a known concentration of another solution. Understanding molarity means you can relate the volume of a solution to the amount of substance it contains, which is essential for titration calculations and for converting between moles and liters of solution.
Balanced Chemical Equations
Chemical reactions are represented by balanced chemical equations, which show the reactants being converted into products. A balanced equation has the same number of each type of atom on both the reactant and product sides, thus obeying the law of conservation of mass. A balanced equation is vital for stoichiometry as it provides the mole-to-mole ratios necessary to perform calculations.

For example, the reaction in the exercise is between sodium hydroxide (NaOH) and hydrochloric acid (HCl) to produce sodium chloride (NaCl) and water (H2O). The balanced equation for this reaction is \( \text{NaOH} + \text{HCl} \rightarrow \text{NaCl} + \text{H}_2\text{O} \), which indicates that one mole of NaOH reacts with one mole of HCl to produce one mole of NaCl and one mole of water. In order to correctly perform calculations around this reaction, ensuring the equation is balanced is crucial.
Mole-to-Mole Ratio
Within a balanced chemical equation, the coefficients offer the mole-to-mole ratios needed to convert between moles of different substances in the reaction. These ratios are critical for stoichiometric calculations because they serve as conversion factors, informing us how many moles of one chemical will react with a certain number of moles of another.

In the titration problem presented, we use the 1:1 mole-to-mole ratio derived from the balanced equation for the reaction between NaOH and HCl. Since the coefficients of NaOH and HCl are both 1, this tells us that for every one mole of NaOH, one mole of HCl is required for a complete reaction, and vice versa. The simplicity of this ratio makes calculations straightforward in this case, but more complex ratios can arise from different balanced equations, which means the mole-to-mole ratio is often the first step in chemical quantity conversions.

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Most popular questions from this chapter

A vitamin C tablet was analyzed to detcrmine whether it did in fact contain, as the manufacturer claimed, \(1.0 \mathrm{~g}\) of the vitamin. A tablet was dissolved in water to form a \(100.00-\mathrm{mL}\). solution, and a \(10.0-\mathrm{mL}\). sample was titrated with iodine (as potassium triiodide). It required \(10.1 \mathrm{~mL}\) of \(0.0521 \mathrm{MI}_{3}\) (aq) to reach the stoichiometric point in the titration, Given that \(1 \mathrm{~mol} \mathrm{I}_{3}^{-}-1 \mathrm{~mol}\) vitamin \(\mathrm{C}\) in the reaction, is the manufacturer's claim correct? The molar mass of vitamin \(\mathrm{C}\) is \(176 \mathrm{~g} \cdot \mathrm{mol}^{-1}\).

A 10.0-mL volume of \(3.0 \mathrm{M} \mathrm{KOH}(\mathrm{aq})\) is transferred to a \(250-\mathrm{mL}\) volumetric flask and diluted to the mark. It was found that \(38.5 \mathrm{~mL}\) of this diluted solution was needed to reach the stoichiometric point in a titration of \(10.0 \mathrm{~mL}\) of a phosphoric acid solution according to the reaction $$ \begin{aligned} 3 \mathrm{KOH}(\mathrm{aq})+& \mathrm{H}_{3} \mathrm{PO}_{4}(\mathrm{aq}) \longrightarrow \\ & \mathrm{K}_{3} \mathrm{PO}_{4}(\mathrm{aq})+3 \mathrm{H}_{2} \mathrm{O}(1) \end{aligned} $$ (a) Calculate the molarity of \(\mathrm{H}_{3} \mathrm{PO}_{4}\) in the solution. (b) What mass of \(\mathrm{H}_{3} \mathrm{PO}_{4}\) is in the initial solution?

When a hydrocarbon burns, water is produced as well as carbon dioxide. (For this reason, clouds of condensed water droplets are often seen coming from automobile exhausts, especially on a cold day.) The density of gasoline is \(0.79 \mathrm{~g} \cdot \mathrm{mL}^{-1}\). Assume gasoline to be represented by octane, \(\mathrm{C}_{8} \mathrm{H}_{18}\), for which the combustion reaction is $$ \begin{aligned} 2 \mathrm{C}_{\mathrm{g}} \mathrm{H}_{18}(\mathrm{l}) &+25 \mathrm{O}_{2}(\mathrm{~g}) \longrightarrow \\ & 16 \mathrm{CO}_{2}(\mathrm{~g})+18 \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \end{aligned} $$ Calculate the mass of water produced from the combustion of \(1.0 \mathrm{~L}\) of gasoline.

Impure phosphoric acid for use in the manufacture of fertilizers is produced by the reaction of sulfuric acid on phosphate rock, of which a principal component is \(\mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)\). The reaction is $$ \begin{aligned} \mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}(\mathrm{~s}) &+3 \mathrm{H}_{2} \mathrm{SO}_{4}(\mathrm{aq}) \longrightarrow \\ & 3 \mathrm{CaSO}_{4}(\mathrm{~s})+2 \mathrm{H}_{3} \mathrm{PO}_{4}(\mathrm{aq}) \end{aligned} $$ (a) How many moles of \(\mathrm{H}_{3} \mathrm{PO}_{4}\) can be produced from the reaction of \(200 \mathrm{~kg}\) of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) ? (b) Determine the mass of calcium sulfate that is produced as a by. product of the reaction of \(200 \mathrm{~mol} \mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}\).

A 15.00-mL. sample of oxalic acid, \(\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\) (with two acidic protons), was titrated to the stoichiometric point with \(17.02 \mathrm{~mL}\) of \(0.288 \mathrm{M} \mathrm{NaOH}(a q)\). (a) What is the molarity of the oxalic acid? (b) Dctermine the mass of oxalic acid in the solution.

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