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Chapter 16: Problem 45

Calculate \(\left[\mathrm{OH}^{-}\right]\) and \(\mathrm{pH}\) for \((\mathbf{a}) 1.5 \times 10^{-3} \mathrm{M} \mathrm{Sr}(\mathrm{OH})_{2},\) (b) \(2.250 \mathrm{~g}\) of \(\mathrm{LiOH}\) in \(250.0 \mathrm{~mL}\) of solution, (c) 1.00 \(\mathrm{mL}\) of \(0.175 \mathrm{M} \mathrm{NaOH}\) diluted to \(2.00 \mathrm{~L},(\mathbf{d})\) a solution formed by adding \(5.00 \mathrm{~mL}\) of \(0.105 \mathrm{M} \mathrm{KOH}\) to \(15.0 \mathrm{~mL}\) of \(9.5 \times 10^{-2} \mathrm{MCa}(\mathrm{OH})_{2}\)

Short Answer

Expert verified
(a) [OH鈦籡 = 0.003 M, pH = 11.48; (b) [OH鈦籡 = 0.3759 M, pH = 13.58; (c) [OH鈦籡 = 8.75 脳 10鈦烩伒 M, pH = 9.94; (d) [OH鈦籡 = 0.0984 M, pH = 12.99.

Step by step solution

01

Identify Ionization in Sr(OH)鈧�

For part (a), identify that one mole of Sr(OH)鈧� produces two moles of OH鈦�. Therefore, the concentration of OH鈦� ions is twice that of the Sr(OH)鈧� concentration.
02

Calculate [OH鈦籡 in Sr(OH)鈧� Solution

Since \[ \text{Sr(OH)}_2 \rightarrow \text{Sr}^{2+} + 2 \text{OH}^- \\left[\text{OH}^鈭抃right] = 2 \times \left[\text{Sr(OH)}_2\right] = 2 \times 1.5 \times 10^{-3} \mathrm{M} = 3.0 \times 10^{-3} \mathrm{M} \\]
03

Calculate pH from OH鈦� Concentration for Sr(OH)鈧�

Use the relation \[ \text{pOH} = -\log[\text{OH}^-] \\text{pH} = 14 - \text{pOH} \\text{pOH} = -\log(3.0 \times 10^{-3}) = 2.52 \\text{pH} = 14 - 2.52 = 11.48 \ \]
04

Convert Mass of LiOH to Moles

For part (b), first convert the mass of LiOH into moles using its molar mass (23.95 g/mol), \[ \text{moles of LiOH} = \frac{2.250 \text{ g}}{23.95 \text{ g/mol}} = 0.09397 \text{ mol} \]
05

Determine LiOH Solution Concentration and OH鈦�

Find the molarity of the LiOH solution, then recognize that it dissociates completely into OH鈦� ions. \[ \text{Molarity of LiOH} = \frac{0.09397 \text{ mol}}{0.250 \text{ L}} = 0.37588 \text{ M} \\left[\text{OH}^鈭抃right] = 0.37588 \text{ M} \]
06

Calculate pH for LiOH Solution

Use the OH鈦� concentration to find pH. \[ \text{pOH} = -\log(0.37588) = 0.425 \\text{pH} = 14 - 0.425 = 13.575 \ \]
07

Calculate New Concentration of NaOH Solution

For part (c), apply dilution based on initial and final volumes.Use the formula \[ C_1V_1 = C_2V_2 \] where \[ C_1 = 0.175 \text{ M} \], \[ V_1 = 0.001 \text{ L} \], and \[ V_2 = 2.00 \text{ L} \].\[ \left[\text{OH}^鈭抃right] = C_2 = \frac{0.175 \times 0.001}{2.00} = 8.75 \times 10^{-5} \text{ M} \]
08

Calculate pH for Diluted NaOH Solution

Calculate pH using the derived OH鈦� concentration.\[ \text{pOH} = -\log(8.75 \times 10^{-5}) = 4.06 \\text{pH} = 14 - 4.06 = 9.94 \ \]
09

Calculate Final [OH鈦籡 for KOH and Ca(OH)鈧� Mixture

For part (d), first calculate moles of OH鈦� from each base and total volume to find combined OH鈦� concentration. \[\text{KOH:} \ \left[\text{OH}^鈭抃right] = 0.105 \text{ M},\ \ \text{Ca(OH)}_2: \ \left[\text{OH}^鈭抃right] = 2 \times 9.5 \times 10^{-2} \ \text{M}\] Calculate moles for each and sum: \[\left(0.105 \times 0.005\right) + \left(2 \times 9.5 \times 10^{-2} \times 0.015\right)= 0.001968 \text{ mol} \ \text{New Total Volume} = 5 \text{ mL} + 15 \text{ mL} = 20 \text{ mL}= 0.020 \text{ L}\ \left[\text{OH}^鈭抃right] = \frac{0.001968}{0.020} = 0.0984 \text{ M} \]
10

Calculate pH for Mixed Solution

Calculate the pH using the final OH鈦� concentration.\[ \text{pOH} = -\log(0.0984) = 1.007 \\text{pH} = 14 - 1.007 = 12.993 \]

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Key Concepts

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

Ion Concentration Calculations
Ion concentration calculation is fundamental when working with chemical substances, especially bases that release hydroxide ions (OH鈦�) in a solution. Determining the concentration of these ions helps us understand the solution's properties. For example, with Sr(OH)鈧�, each formula unit releases two OH鈦� ions upon dissociation. To calculate the ion concentration, you multiply the given molarity of Sr(OH)鈧� by two, since two moles of OH鈦� are produced per mole of Sr(OH)鈧�.

When dealing with solid compounds like LiOH, transformation from mass to moles is required before calculating ion concentration. The proportional relationship:
  • Mass of LiOH 鈫� Moles via molar mass
  • Moles 鈫� Molarity in solution via volume
helps convert the initial mass into a concentration in molarity (moles per liter). Once the molarity of the compound in solution is known, it equals the OH鈦� concentration because LiOH dissociates fully into hydroxide ions.

This type of ion concentration calculation is vital for pH determination and understanding the reactivity and strength of the base.
Alkaline Solutions
Alkaline or basic solutions contain more hydroxide ions than hydrogen ions. The strength of these solutions is commonly expressed in terms of pH, a logarithmic scale where 7 is neutral, less than 7 is acidic, and more than 7 is basic. Strong alkaline solutions, such as those derived from LiOH and NaOH, have high pH values because they dissociate completely in water, releasing OH鈦� ions.

To find the pH of an alkaline solution, we first calculate the pOH using the formula:

oindent\( \text{pOH} = -\log[\text{OH}^-] \)

Then, we use:

oindent\(\text{pH} = 14 - \text{pOH} \)

The manipulation of these formulas shows how dominant the hydroxide ion concentration affects the resulting pH. A higher OH鈦� concentration results in a lower pOH and thus a higher pH, indicating a more alkaline solution.

Understanding these relationships helps when predicting how changes in solution, such as dilution, will impact overall alkalinity.
Chemical Solution Dilution
Dilution involves adding more solvent to a solution, decreasing the concentration of solutes. In chemical solutions, this process involves using the concentration equation:

\( C_1V_1 = C_2V_2 \)

where \(C_1\) and \(V_1\) are the initial concentration and volume, and \(C_2\) and \(V_2\) are the final concentration and volume, respectively.

This fundamental principle of dilution is crucial, especially for experiments or reactions where precise concentrations are needed. For example, diluting a 0.175 M NaOH solution by a significant volume reduces its concentration dramatically, affecting both the chemical reactivity and the pH.

The diluted solution's hydroxide ion concentration can still be calculated with accuracy using the dilution formula, allowing us to determine how the pH value鈥攁 critical measure of concentration鈥攃hanges upon dilution. Therefore, understanding and applying dilution principles ensures proper pH and concentration determination in various chemical processes.

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Most popular questions from this chapter

(a) Which of the following is the stronger Bronsted-Lowry acid, \(\mathrm{HClO}_{3}\) or \(\mathrm{HClO}_{2} ?\) (b) Which is the stronger BronstedLowry base, \(\mathrm{HS}^{-}\) or \(\mathrm{HSO}_{4}^{-}\) ?

The amino acid glycine \(\left(\mathrm{H}_{2} \mathrm{~N}-\mathrm{CH}_{2}-\mathrm{COOH}\right)\) can participate in the following equilibria in water: \(\mathrm{H}_{2} \mathrm{~N}-\mathrm{CH}_{2}-\mathrm{COOH}+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons\) $$ \mathrm{H}_{2} \mathrm{~N}-\mathrm{CH}_{2}-\mathrm{COO}^{-}+\mathrm{H}_{3} \mathrm{O}^{+} \quad K_{\mathrm{a}}=4.3 \times 10^{-3} $$$$ \begin{aligned} \mathrm{H}_{2} \mathrm{~N}-\mathrm{CH}_{2}-\mathrm{COOH}+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \\ &{ }^{+} \mathrm{H}_{3} \mathrm{~N}-\mathrm{CH}_{2}-\mathrm{COOH}+\mathrm{OH}^{-} \quad K_{\mathrm{b}}=6.0 \times 10^{-5} \end{aligned} $$ (a) Use the values of \(K_{a}\) and \(K_{b}\) to estimate the equilibrium constant for the intramolecular proton transfer to form a zwitterion: $$ \mathrm{H}_{2} \mathrm{~N}-\mathrm{CH}_{2}-\mathrm{COOH} \rightleftharpoons{ }^{+} \mathrm{H}_{3} \mathrm{~N}-\mathrm{CH}_{2}-\mathrm{COO}^{-} $$ (b) What is the pH of a \(0.050 \mathrm{M}\) aqueous solution of glycine? (c) What would be the predominant form of glycine in a solution with \(\mathrm{pH} 13\) ? With \(\mathrm{pH} 1\) ?

Ammonia, \(\mathrm{NH}_{3}\), acts as an Arrhenius base, a Br酶nsted-Lowry base, and a Lewis base, in aqueous solution. Write out the reaction \(\mathrm{NH}_{3}\) undergoes with water and explain what properties of ammonia correspond to each of the three definitions of "base."

Calculate the concentration of an aqueous solution of \(\mathrm{Ca}(\mathrm{OH})_{2}\) that has a pH of 10.05 .

A particular sample of vinegar has a pH of 2.20 . If acetic acid is the only acid that vinegar contains \(\left(K_{a}=1.8 \times 10^{-5}\right)\), calculate the concentration of acetic acid in the vinegar.
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