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Chapter 1: Problem 13

\(10.0 \mathrm{~mL}\) of \(0.5 \mathrm{M} \mathrm{HC} 1\) was used to titrate \(5.0 \mathrm{~mL}\) of \(\mathrm{NaOH}\) to its equivalence point. The molarity of the \(\mathrm{NaOH}\) is: a. \(0.5\) b. \(2.0\) c. \(1.0\) d. \(0.1\) e. \(0.2\)

Short Answer

Expert verified
The molarity of the \(\mathrm{NaOH}\) is 1.0 M.

Step by step solution

01

- Write the Chemical Equation

The balanced chemical equation for the reaction between HCl and NaOH is: Discuss the beginning of the reaction: \[ \mathrm{HCl} (aq) + \mathrm{NaOH} (aq) \rightarrow \mathrm{NaCl} (aq) + \mathrm{H_2O} (l) \].
02

- Determine Moles of HCl

Use the formula for molarity: \[ M = \frac{n}{V} \] where \(M\) is molarity, \(n\) is the number of moles, and \(V\) is the volume in liters. Given \(M = 0.5 \mathrm{M}\) and \(V = 10.0 \mathrm{~mL} = 0.0100 \mathrm{L}\), calculate the moles of \(\mathrm{HCl}\):\[ n_{\mathrm{HCl}} = 0.5 \mathrm{M} \times 0.0100 \mathrm{L} = 0.005 \mathrm{mol} \]
03

- Relate Moles of HCl to Moles of NaOH

At the equivalence point of the titration, the moles of \(\mathrm{HCl}\) equal the moles of \(\mathrm{NaOH}\): \[ n_{\mathrm{NaOH}} = n_{\mathrm{HCl}} = 0.005 \mathrm{mol} \].
04

- Calculate Molarity of NaOH

Using the formula for molarity again, where \(V = 5.0 \mathrm{~mL} = 0.0050 \mathrm{L}\), Calculate the molarity of \(\mathrm{NaOH}\): \[ M_{\mathrm{NaOH}} = \frac{n}{V} = \frac{0.005 \mathrm{mol}}{0.0050 \mathrm{L}} = 1.0 \mathrm{M} \]

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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 measure of the concentration of a solute in a solution. It is calculated using the formula: \( M = \frac{n}{V} \) where \(M\) is molarity, \(n\) is the number of moles of the solute, and \(V\) is the volume of the solution in liters. To find molarity, simply divide the number of moles by the volume in liters. For example, if you have 0.005 moles of HCl dissolved in 0.0100 liters of solution, the molarity would be: \( M = \frac{0.005 \text{ mol}}{0.0100 \text{ L}} = 0.5 \text{ M} \).
Acid-Base Reaction
In an acid-base reaction, an acid (HCl in this case) reacts with a base (NaOH) to produce water and a salt. The balanced chemical equation is: \[ \mathrm{HCl} (aq) + \mathrm{NaOH} (aq) \rightarrow \mathrm{NaCl} (aq) + \mathrm{H_2O} (l) \] This reaction shows that one molecule of HCl reacts with one molecule of NaOH to produce one molecule of NaCl and one molecule of water. This simple 1:1 ratio is critical because it means that the moles of HCl and NaOH will be equal at the equivalence point.
Equivalence Point
The equivalence point in a titration is the moment when the quantity of titrant added is exactly enough to completely react with the analyte in the solution. For an acid-base reaction like HCl and NaOH, this is when the moles of HCl added are equal to the moles of NaOH present in the solution. At this point, there are no excess HCl or NaOH molecules; they have completely neutralized each other.
Stoichiometry
Stoichiometry is the part of chemistry that deals with the amounts of reactants and products in chemical reactions. It is based on the balanced chemical equation and uses the mole ratio to relate the quantities of reactants and products. In the case of the reaction between HCl and NaOH: \[\mathrm{HCl} (aq) + \mathrm{NaOH} (aq) \rightarrow \mathrm{NaCl} (aq) + \mathrm{H_2O} (l)\] The coefficients (1:1:1:1) show that one mole of HCl reacts with one mole of NaOH to make one mole of NaCl and one mole of water. Using this 1:1 ratio, if you know the moles of one reactant (HCl), you can directly determine the moles of the other (NaOH). Therefore, if you have 0.005 moles of HCl, you also have 0.005 moles of NaOH at the equivalence point.

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