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Chapter 16: Problem 45

Calculate \(\left[\mathrm{OH}^{-}\right]\) and \(\mathrm{pH}\) for (a) $1.5 \times 10^{-3} \mathrm{MSr}(\mathrm{OH})_{2},\( (b) \)2.250 \mathrm{~g}$ of \(\mathrm{LiOH}\) in \(250.0 \mathrm{~mL}\) of solution, \((\mathbf{c}) 1.00\) \(\mathrm{mL}\) 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⁻] = 3.0 x 10⁻³ M; pH = 11.48 (b) [OH⁻] = 0.392 M; pH = 13.593 (c) [OH⁻] = 8.75 x 10⁻⁵ M; pH = 9.94 (d) [OH⁻] = 0.16875 M; pH = 13.227

Step by step solution

01

(a) Finding [OH⁻] for 1.5 x 10⁻³ M Sr(OH)₂

We are given the molar concentration of Sr(OH)₂, and we know that one mole of Sr(OH)₂ dissociates into two moles of OH⁻ ions. So, the [OH⁻] can be calculated as follows: [OH⁻] = 2 * (1.5 x 10⁻³) = 3.0 x 10⁻³ M
02

(a) Finding pH for 1.5 x 10⁻³ M Sr(OH)₂

First, we need to find the pOH: pOH = -log(3.0 x 10⁻³) = 2.52 Now, we can calculate the pH using the relationship: pH = 14 - pOH = 14 - 2.52 = 11.48
03

(b) Finding [OH⁻] for 2.250 g of LiOH in 250.0 mL of solution

First, we need to calculate the moles of LiOH: Moles of LiOH = mass / molar mass = 2.250 g / (6.94 g/mol + 15.999 g/mol) = 2.250 g / 22.939 g/mol = 0.0981 mol Next, we need to find the molar concentration: [LiOH] = moles / volume = 0.0981 mol / 0.250 L = 0.392 M Since one mole of LiOH dissociates into one mole of OH⁻ ions, the [OH⁻] is equal to the [LiOH]: [OH⁻] = 0.392 M
04

(b) Finding the pH for 2.250 g of LiOH in 250.0 mL of solution

First, we need to find the pOH: pOH = -log(0.392) = 0.407 Now, we can calculate the pH using the relationship: pH = 14 - pOH = 14 - 0.407 = 13.593
05

(c) Finding [OH⁻] for 1.00 mL of 0.175 M NaOH diluted to 2.00 L

First, we need to find the moles of NaOH: Moles of NaOH = volume * concentration = 0.001 L * 0.175 M = 0.000175 mol Next, we need to find the molar concentration after dilution: [OH⁻] = moles / volume = 0.000175 mol / 2.00 L = 8.75 x 10⁻⁵ M
06

(c) Finding the pH for 1.00 mL of 0.175 M NaOH diluted to 2.00 L

First, we need to find the pOH: pOH = -log(8.75 x 10⁻⁵) = 4.06 Now, we can calculate the pH using the relationship: pH = 14 - pOH = 14 - 4.06 = 9.94
07

(d) Finding [OH⁻] for a solution formed by adding 5.00 mL of 0.105 M KOH to 15.0 mL of 9.5 x 10⁻² M Ca(OH)₂

First, we need to find the moles of OH⁻ ions from each compound: Moles OH⁻ from KOH = volume * concentration = 0.005 L * 0.105 M = 5.25 x 10⁻⁴ mol Moles OH⁻ from Ca(OH)₂ = volume * concentration * 2 = 0.015 L * 9.5 x 10⁻² M * 2 = 2.85 x 10⁻³ mol Now, we need to find the total moles of OH⁻ ions and calculate the molar concentration: Total moles OH⁻ = 5.25 x 10⁻⁴ mol + 2.85 x 10⁻³ mol = 3.375 x 10⁻³ mol [OH⁻] = total moles OH⁻ / total volume = 3.375 x 10⁻³ mol / 0.020 L = 0.16875 M
08

(d) Finding the pH for a solution formed by adding 5.00 mL of 0.105 M KOH to 15.0 mL of 9.5 x 10⁻² M Ca(OH)₂

First, we need to find the pOH: pOH = -log(0.16875) = 0.773 Now, we can calculate the pH using the relationship: pH = 14 - pOH = 14 - 0.773 = 13.227

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Most popular questions from this chapter

At \(50^{\circ} \mathrm{C}\), the ion-product constant for $\mathrm{H}_{2} \mathrm{O}\( has the value \)K_{w}=5.48 \times 10^{-14} \cdot(\mathbf{a})$ What is the \(\mathrm{pH}\) of pure water at \(50^{\circ} \mathrm{C} ?\) (b) Based on the change in \(K_{w}\) with temperature, predict whether \(\Delta H\) is positive, negative, or zero for the autoionization reaction of water: $$ 2 \mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{H}_{3} \mathrm{O}^{+}(a q)+\mathrm{OH}^{-}(a q) $$

Addition of phenolphthalein to an unknown colorless solution does not cause a color change. The addition of bromthymol blue to the same solution leads to a yellow color. (a) Is the solution acidic, neutral, or basic? (b) Which of the following can you establish about the solution: (i) A minimum \(\mathrm{pH}\), (ii) A maximum \(\mathrm{pH}\), or (iii) A specific range of pH values? (c) What other indicator or indicators would you want to use to determine the \(\mathrm{pH}\) of the solution more precisely?

A \(0.25 M\) solution of a salt NaA has \(\mathrm{pH}=9.29 .\) What is the value of \(K_{a}\) for the parent acid HA?

Ammonia, \(\mathrm{NH}_{3}\), acts as an Arrhenius base, a Brønsted-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 percent ionization of hydrazoic acid \(\left(\mathrm{HN}_{3}\right)\) in solutions of each of the following concentrations \(\left(K_{a}\right.\) is (c) \(0.0400 \mathrm{M}\). given in Appendix $\mathrm{D}):(\mathbf{a}) 0.400 \mathrm{M},(\mathbf{b}) 0.100 \mathrm{M}$
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