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Chapter 16: Problem 25

Estimate the \(\mathrm{pH}\) that results when the following two solutions are mixed. (a) \(50 \mathrm{~mL}\) of \(0.1 \mathrm{M} \mathrm{HCl}\) and \(50 \mathrm{~mL}\) of \(0.2 \mathrm{M} \mathrm{NaOH}\) (b) \(100 \mathrm{~mL}\) of \(0.1 \mathrm{M} \mathrm{HCl}\) and \(50 \mathrm{~mL}\) of \(0.2 \mathrm{M} \mathrm{NaOH}\) (c) \(150 \mathrm{~mL}\) of \(0.1 \mathrm{MHCl}\) and \(50 \mathrm{~mL}\) of \(0.2 \mathrm{M} \mathrm{Ba}(\mathrm{OH})_{2}\) (d) \(200 \mathrm{~mL}\) of \(0.1 \mathrm{M} \mathrm{HCl}\) and \(50 \mathrm{~mL}\) of \(0.2 \mathrm{M} \mathrm{Ba}(\mathrm{OH})_{2}\)

Short Answer

Expert verified
(a) pH = 12.7, (b) pH = 7.0, (c) pH = 12.4, (d) pH = 7.0

Step by step solution

01

Calculate moles of HCl and NaOH in (a)

For (a), first calculate the moles of HCl: \[ n_{ ext{HCl}} = C imes V = 0.1 ext{ M} imes 50 ext{ mL} imes rac{1 ext{ L}}{1000 ext{ mL}} = 0.005 ext{ moles} \]Calculate the moles of NaOH:\[ n_{ ext{NaOH}} = 0.2 ext{ M} imes 50 ext{ mL} imes rac{1 ext{ L}}{1000 ext{ mL}} = 0.01 ext{ moles} \]Since NaOH has more moles, it will neutralize HCl, and extra NaOH will remain.
02

Determine resulting pH for (a)

The moles of NaOH left after neutralizing HCl in (a):\[ n_{ ext{NaOH remaining}} = 0.01 - 0.005 = 0.005 ext{ moles} \]The total volume of the mixed solution is:\[ V_{ ext{total}} = 50 ext{ mL} + 50 ext{ mL} = 100 ext{ mL} = 0.1 ext{ L} \]The concentration of OH鈦� in the solution:\[ [ ext{OH}^-] = rac{0.005}{0.1} = 0.05 ext{ M} \]Calculate pOH and pH:\[ ext{pOH} = - ext{log}(0.05) = 1.30 \]\[ ext{pH} = 14 - 1.30 = 12.7 \]
03

Calculate moles and resulting pH for (b)

For (b), calculate the moles of HCl and NaOH:Moles of HCl:\[ n_{ ext{HCl}} = 0.1 ext{ M} imes 100 ext{ mL} imes rac{1 ext{ L}}{1000 ext{ mL}} = 0.01 ext{ moles} \]Moles of NaOH:\[ n_{ ext{NaOH}} = 0.2 ext{ M} imes 50 ext{ mL} imes rac{1 ext{ L}}{1000 ext{ mL}} = 0.01 ext{ moles} \]All the HCl and NaOH will neutralize each other completely, so the resulting pH is 7 as the solution is neutral.
04

Calculate moles for (c) and determine excess component

For (c), calculate the moles of HCl and Ba(OH)鈧�:Moles of HCl:\[ n_{ ext{HCl}} = 0.1 ext{ M} imes 150 ext{ mL} imes rac{1 ext{ L}}{1000 ext{ mL}} = 0.015 ext{ moles} \]Ba(OH)鈧� provides 2 moles of OH鈦� per mole:\[ n_{ ext{OH}^-} = 0.2 ext{ M} imes 50 ext{ mL} imes rac{1 ext{ L}}{1000 ext{ mL}} imes 2 = 0.02 ext{ moles} \]Excess OH鈦�:\[ n_{ ext{OH}^- remaining} = 0.02 - 0.015 = 0.005 ext{ moles} \]Calculate pH using the total volume:\[ V_{ ext{total}} = 150 ext{ mL} + 50 ext{ mL} = 200 ext{ mL} = 0.2 ext{ L} \]\[ [ ext{OH}^-] = rac{0.005}{0.2} = 0.025 ext{ M} \]\[ ext{pOH} = - ext{log}(0.025) = 1.60 \]\[ ext{pH} = 14 - 1.60 = 12.4 \]
05

Calculate moles for (d) and determine solution pH

For (d), calculate the moles of HCl and Ba(OH)鈧�:Moles of HCl:\[ n_{ ext{HCl}} = 0.1 ext{ M} imes 200 ext{ mL} imes rac{1 ext{ L}}{1000 ext{ mL}} = 0.02 ext{ moles} \]Wall (OH鈦�):\[ n_{ ext{OH}^-} = 0.2 ext{ M} imes 50 ext{ mL} imes rac{1 ext{ L}}{1000 ext{ mL}} imes 2 = 0.02 ext{ moles} \]Both neutralize each other completely, resulting in a neutral pH.So the pH is:\[ ext{pH} = 7.0 \]

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Key Concepts

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

Acid-Base Neutralization
Acid-base neutralization is a fundamental concept in chemistry where an acid and a base react together to form water and a salt. This reaction is important in understanding how to achieve a neutral solution, typically having a pH of 7.
When a strong acid, such as hydrochloric acid (HCl), meets a strong base like sodium hydroxide (NaOH), they neutralize each other's reactive properties. This means the hydrogen ions (\(H^+\)) from the acid combine with the hydroxide ions (\(OH^-\)) from the base to build water (\(H_2O\)).
For instance, when equal moles of HCl and NaOH react, the result is complete neutralization, and the solution becomes neutral. However, if there is an excess of one over the other, the pH shifts towards acidic or basic depending on which is in excess.
  • HCl + NaOH 鈫� NaCl + H鈧侽
  • H鈦� combines with OH鈦� to produce H鈧侽
Understanding neutralization is crucial for calculating the balance between acids and bases in various reactions. It allows for the estimation of the resultant pH when mixing different volumes and concentrations of acids and bases.
Molarity and Concentration
Molarity, a measure of concentration, defines the number of moles of solute per liter of solution. It's a fundamental aspect of solution chemistry because it specifies how much solute is dissolved in a given volume. Molarity is expressed in \(M\), representing moles per liter (mol/L).
The molarity helps in determining how solutions will react when combined. For example, knowing the molarity of an acid and a base helps predict how much of each is needed to reach neutralization. In calculations, you may convert milliliters to liters when using molarity, ensuring units are consistent.
In our exercise, we work with solutions like \(0.1 M\) HCl and \(0.2 M\) NaOH. Their molarities aid in calculating how many moles of each component present in the given solution volume will participate in the neutralization reaction.
  • Molarity (\(C\)) = Moles of solute (\(n\)) / Volume of solution (\(V\)) in Liters
  • To convert mL to L: multiply by 0.001 or divide by 1000
Accurate molarity calculation is pivotal for predicting results when acids and bases are mixed, affecting the resulting solution's pH.
Solution Mixing
Solution mixing refers to combining two or more solutions to achieve a desired chemical reaction or solution composition. It is a common practice in labs, especially when trying to achieve specific pH values or concentrations.
When solutions like acids and bases are mixed, their volumes combine, which needs accounting to determine the concentration after mixing. Volume changes impact the concentration of ions in the final solution, significantly determining the resulting pH.
  • Total Volume = Volume of solution 1 + Volume of solution 2
  • Adjust concentrations based on the new total volume
Consider the exercise where different volumes of HCl and NaOH solutions are mixed. By knowing the initial molarity and volume, you can predict the final concentration of ions in the combined solution, facilitating pH calculation. Classically, applying the principles of molarity and neutralization can accurately determine the resultant pH when these solutions undergo mixing.

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Most popular questions from this chapter

A Saccharin is an artificial sweetener that is also a weak acid. It has the formula \(\mathrm{C}_{7} \mathrm{H}_{5} \mathrm{NSO}_{3},\) and its \(\mathrm{p} K_{\mathrm{a}}\) is \(11.68 .\) A 12 -oz \((350-\mathrm{mL})\) can of diet cola contains \(3.0 \mathrm{mg} \mathrm{sac}-\) charin and has a pH of \(4.50 .\) What are the equilibrium molar concentrations of saccharin and the saccharide ion?

Phenolphthalein is a commonly used indicator that is colorless in the acidic form \((\mathrm{pH}<8.3)\) and pink in the base form \((\mathrm{pH}>10.0)\). It is a weak acid with a \(\mathrm{pK}_{\mathrm{a}}\) of 8.7 . What fraction is in the acid form when the acid color is apparent? What fraction is in the base form when the base color is apparent?

A Calculate the minimum concentrations of acetic acid and sodium acetate that are needed to prepare \(100 \mathrm{~mL}\) of a pH 4.50 buffer whose \(\mathrm{pH}\) will not change by more than 0.05 unit if \(1.00 \mathrm{~mL}\) of \(0.100 \mathrm{M}\) strong acid or strong base is added.

A monoprotic organic acid that has a molar mass of \(176.1 \mathrm{~g} / \mathrm{mol}\) is synthesized. Unfortunately, the acid produced is not completely pure. In addition, it is not soluble in water. A chemist weighs a \(1.8451-\mathrm{g}\) sample of the impure acid and adds it to \(100.0 \mathrm{~mL}\) of \(0.1050 \mathrm{M} \mathrm{NaOH}\). The acid is soluble in the \(\mathrm{NaOH}\) solution and reacts to consume most of the \(\mathrm{NaOH}\). The amount of excess \(\mathrm{NaOH}\) is determined by titration: It takes \(3.28 \mathrm{~mL}\) of \(0.0970 \mathrm{M} \mathrm{HCl}\) to neutralize the excess \(\mathrm{NaOH}\). What is the purity of the original acid, in percent?

A Concentrated hydrochloric acid is \(38 \% \mathrm{HCl}\) by weight and has a density of \(1.19 \mathrm{~g} / \mathrm{mL}\). A solution is prepared by measuring \(83 \mathrm{~mL}\) of the concentrated \(\mathrm{HCl}\), adding it to water, and diluting to \(1.00 \mathrm{~L}\) (a) Calculate the approximate molarity of this solution from the volume, percentage composition, and density. (b) The exact concentration is determined by titration. \(A\) \(25.00-\mathrm{mL}\) portion of the solution is titrated with \(1.04 M \mathrm{NaOH} .\) Phenolphthalein changes color after adding \(23.88 \mathrm{~mL}\) of the base. What is the concentration of the HCl solution? How does the approximate concentration calculated in part a compare with the exact concentration?
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