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Chapter 15: Problem 88

What quantity (moles) of \(\mathrm{HCl}(g)\) must be added to \(1.0 \mathrm{~L}\) of 2.0 \(M \mathrm{NaOH}\) to achieve a pH of \(0.00 ?\) (Neglect any volume changes.)

Short Answer

Expert verified
To achieve a pH of \(0.00\) in \(1.0 \mathrm{~L}\) of 2.0 \(M \mathrm{NaOH}\), you need to add \(3.0 \mathrm{~moles}\) of \(\mathrm{HCl}(g)\).

Step by step solution

01

Calculate initial moles of \(\mathrm{NaOH}\)

The initial moles of \(\mathrm{NaOH}\) can be calculated using the initial volume and molarity. Initial moles of \(\mathrm{NaOH}\) = Molarity 脳 Volume Initial moles of \(\mathrm{NaOH}\) = \(2.0M \times 1.0L = 2.0\) moles
02

Calculate the final concentration of hydrogen ions \(([H^+])\)

Since we want a pH of \(0.00\), we will find out the final concentration of hydrogen ions using the pH formula: \(pH = -\log[H^+]\) We need to rearrange the formula to solve for \([H^+]\): \([H^+] = 10^{-pH}\) Hence, \([H^+] = 10^{-0} = 1 \mathrm{M}\)
03

Find moles of \(H^+\) ions required to achieve the desired pH

Now we will find the moles of \(H^+\) ions required to achieve the desired pH using the final concentration of \(H^+\) ions and the final volume of the solution: Moles of \(H^+\) ions = Concentration 脳 Volume Moles of \(H^+\) ions = \(1 \mathrm{M} \times 1.0 \mathrm{L} = 1.0\) moles
04

Determine moles of \(HCl\) required to neutralize the given \(\mathrm{NaOH}\)

In order to neutralize the \(2.0\) moles of \(\mathrm{NaOH}\), we need an equal number of moles of \(\mathrm{HCl}\) (both being strong acid and strong base; they neutralize each other): Neutralization reaction: \(\mathrm{NaOH + HCl \rightarrow NaCl + H_2O}\) Moles of \(\mathrm{HCl}\) required for neutralization = Moles of \(\mathrm{NaOH}\) Moles of \(\mathrm{HCl}\) required for neutralization = \(2.0\) moles
05

Calculate the total moles of \(\mathrm{HCl}\) to be added

Now, we need to add the moles of \(\mathrm{HCl}\) required for neutralization and the moles of \(H^+\) ions required to achieve the desired pH: Total moles of \(\mathrm{HCl}\) = Moles of \(\mathrm{HCl}\) required for neutralization + Moles of \(H^+\) ions required for desired pH Total moles of \(\mathrm{HCl}\) = \(2.0 \mathrm{~moles} + 1.0 \mathrm{~moles} = 3.0 \mathrm{~moles}\) Therefore, \(3.0 \mathrm{~moles}\) of \(\mathrm{HCl}(g)\) must be added to \(1.0 \mathrm{~L}\) of 2.0 \(M \mathrm{NaOH}\) to achieve a pH of \(0.00\).

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Key Concepts

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

pH calculation
pH is a measure of how acidic or basic a solution is. It is calculated using the formula:\[pH = -\log[H^+]\]Where
  • \([H^+]\) is the concentration of hydrogen ions in the solution.
  • pH values less than 7 indicate an acidic solution, a value of 7 is neutral, and values greater than 7 are basic.
To find the concentration of hydrogen ions based on the desired pH, you can rearrange the formula:\[[H^+] = 10^{-pH}\]For example, if you need a pH of 0.00, the hydrogen ion concentration becomes:\[[H^+] = 10^{-0} = 1 \text{ M}\]This means you need a 1 mole per liter concentration of hydrogen ions to achieve this acidity.
molarity and volume
Molarity is a way to express the concentration of a solution in terms of the amount of solute per volume of solution. It's defined as:\[\text{Molarity (M)} = \frac{\text{moles of solute}}{\text{liters of solution}}\]In many chemical calculations, combining molarity with the volume of the solution helps you find out how many moles of a solute you have. For instance, if you have a solution with a molarity of \(2.0 \text{ M NaOH}\) and a volume of \(1.0 \text{ L}\), you can calculate the initial moles of sodium hydroxide (NaOH) involved:\[\text{Moles of } NaOH = 2.0 \text{ M} \times 1.0 \text{ L} = 2.0 \text{ moles}\]This principle is essential for planning reactions, such as ensuring enough neutralization can occur when an acid is introduced.
neutralization reaction
A neutralization reaction occurs when an acid and a base react to form water and salt, effectively cancelling each other's reactive properties. The reaction can be represented as:\[\text{Base (e.g., NaOH) + Acid (e.g., HCl) } \rightarrow \text{ Salt (e.g., NaCl) + Water } (H_2O)\]In our example, \(\text{HCl}(g)\) is added to \(\text{NaOH}\), a strong acid reacting with a strong base, producing sodium chloride and water.
  • The goal is to reach a state where there is no excess of hydrogen ions \([H^+]\) or hydroxide ions \([OH^-]\) left in the solution.
  • The number of moles of hydrochloric acid required is equal to the moles of sodium hydroxide, due to their 1:1 molar ratio in the balanced equation.
  • In our specific problem, \(2.0 \text{ moles of } HCl\) are needed to neutralize \(2.0 \text{ moles of } NaOH\).
After neutralizing NaOH, additional HCl is added to achieve the desired pH, illustrating the thorough application of stoichiometric principles to achieve specific chemical properties in a solution.

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Most popular questions from this chapter

Consider \(1.0 \mathrm{~L}\) of a solution that is \(0.85 \mathrm{M} \mathrm{HOC}_{6} \mathrm{H}_{5}\) and \(0.80 \mathrm{M} \mathrm{NaOC}_{6} \mathrm{H}_{5} .\left(K_{\mathrm{a}}\right.\) for \(\left.\mathrm{HOC}_{6} \mathrm{H}_{5}=1.6 \times 10^{-10} \cdot\right)\) a. Calculate the \(\mathrm{pH}\) of this solution. b. Calculate the \(\mathrm{pH}\) after \(0.10 \mathrm{~mole}\) of \(\mathrm{HCl}\) has been added to the original solution. Assume no volume change on addition of \(\mathrm{HCl}\). c. Calculate the \(\mathrm{pH}\) after \(0.20 \mathrm{~mole}\) of \(\mathrm{NaOH}\) has been added to the original buffer solution. Assume no volume change on addition of \(\mathrm{NaOH}\).

Which of the following mixtures would result in buffered solutions when \(1.0 \mathrm{~L}\) of each of the two solutions are mixed? a. \(0.1 \mathrm{M} \mathrm{KOH}\) and \(0.1 \mathrm{M} \mathrm{CH}_{3} \mathrm{NH}_{3} \mathrm{Cl}\) b. \(0.1 \mathrm{M} \mathrm{KOH}\) and \(0.2 \mathrm{M} \mathrm{CH}_{3} \mathrm{NH}_{2}\) c. \(0.2 \mathrm{M} \mathrm{KOH}\) and \(0.1 \mathrm{M} \mathrm{CH}_{3} \mathrm{NH}_{3} \mathrm{Cl}\) d. \(0.1 \mathrm{M} \mathrm{KOH}\) and \(0.2 \mathrm{M} \mathrm{CH}_{3} \mathrm{NH}_{3} \mathrm{Cl}\)

Calculate the \(\mathrm{pH}\) after \(0.010\) mole of gaseous \(\mathrm{HCl}\) is added to \(250.0 \mathrm{~mL}\) of each of the following buffered solutions. a. \(0.050 \mathrm{M} \mathrm{NH}_{3} / 0.15 \mathrm{M} \mathrm{NH}_{4} \mathrm{Cl}\) b. \(0.50 \mathrm{M} \mathrm{NH}_{3} / 1.50 \mathrm{M} \mathrm{NH}_{4} \mathrm{Cl}\) Do the two original buffered solutions differ in their \(\mathrm{pH}\) or their capacity? What advantage is there in having a buffer with a greater capacity?

A buffered solution is made by adding \(50.0 \mathrm{~g} \mathrm{NH}_{4} \mathrm{Cl}\) to \(1.00\) \(\mathrm{L}\) of a \(0.75-M\) solution of \(\mathrm{NH}_{3} .\) Calculate the \(\mathrm{pH}\) of the final solution. (Assume no volume change.)

A few drops of each of the indicators shown in the accompanying table were placed in separate portions of a \(1.0-M\) solution of a weak acid, HX. The results are shown in the last column of the table. What is the approximate \(\mathrm{pH}\) of the solution containing HX? Calculate the approximate value of \(K_{\mathrm{a}}\) for HX. $$ \begin{array}{|lcccc|} \hline \text { Indicator } & \begin{array}{c} \text { Color } \\ \text { of Hin } \end{array} & \begin{array}{c} \text { Color } \\ \text { of In }^{-} \end{array} & \begin{array}{c} \mathrm{pK}_{\mathrm{a}} \\ \text { of Hin } \end{array} & \begin{array}{c} \text { Color of } \\ 1.0 \mathrm{MHX} \end{array} \\ \text { Bromphenol blue } & \text { Yellow } & \text { Blue } & 4.0 & \text { Blue } \\ \text { Bromcresol purple } & \text { Yellow } & \text { Purple } & 6.0 & \text { Yellow } \\ \text { Bromcresol green } & \text { Yellow } & \text { Blue } & 4.8 & \text { Green } \\ \text { Alizarin } & \text { Yellow } & \text { Red } & 6.5 & \text { Yellow } \\\ \hline \end{array} $$
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