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Chapter 16: Problem 46

Calculate \(\left[\mathrm{OH}^{-}\right]\) and \(\mathrm{pH}\) for each of the following strong base solutions: \((\mathbf{a}) 0.182 \mathrm{M} \mathrm{KOH},(\mathbf{b}) 3.165 \mathrm{g}\) of \(\mathrm{KOH}\) in 500.0 mL of solution, ( c ) 10.0 \(\mathrm{mL}\) of 0.0105 \(\mathrm{MCa}(\mathrm{OH})_{2}\) diluted to \(500.0 \mathrm{mL},(\mathbf{d})\) a solution formed by mixing 20.0 \(\mathrm{mL}\) of 0.015 \(M \mathrm{Ba}(\mathrm{OH})_{2}\) with 40.0 \(\mathrm{mL}\) of \(8.2 \times 10^{-3} \mathrm{M} \mathrm{NaOH}.\)

Short Answer

Expert verified
The short answers for each of the strong base solutions are: a) [OH⁻] = 0.182 M and pH ≈ -0.74 b) [OH⁻] = 0.113 M and pH ≈ -0.95 c) [OH⁻] ≈ 4.2 × 10⁻⁴ M and pH ≈ -3.38 d) [OH⁻] ≈ 1.55 × 10⁻² M and pH ≈ -1.81

Step by step solution

01

Find [OH⁻] concentration

Since KOH is a strong base, we know that [OH⁻] = [KOH]. Therefore, [OH⁻] = 0.182 M.
02

Calculate the pH

We can convert the [OH⁻] to pH using the formula: pH = -log[OH⁻] = -log(0.182) ≈ -0.74. #b.# Calculate [OH⁻] and pH for 3.165 g of KOH in 500.0 mL of solution
03

Calculate the molar concentration

First, we need to convert the mass of KOH into moles. moles of KOH = (3.165 g) / (56.11 g/mol) ≈ 0.0564 mol. Now, we can calculate the molar concentration: M = moles / volume = 0.0564 mol / 0.5 L ≈ 0.113 M.
04

Find [OH⁻] concentration

Since KOH is a strong base, [OH⁻] = [KOH]. Therefore, [OH⁻] = 0.113 M.
05

Calculate the pH

We can convert the [OH⁻] to pH using the formula: pH = -log[OH⁻] = -log(0.113) ≈ -0.95. #c.# Calculate [OH⁻] and pH for 10.0 mL of 0.0105 MCa(OH)₂ diluted to 500.0 mL
06

Calculate the final concentration

Since the 10 mL of 0.0105 M solution was diluted, we need to use the dilution formula: M1V1 = M2V2. Solving for M2: M2 = (0.0105 M * 0.010 L) / 0.5 L ≈ 2.1 × 10⁻⁴ M.
07

Find [OH⁻] concentration

Since Ca(OH)₂ produces two moles of OH⁻ ions for every mole of Ca(OH)₂, our [OH⁻] = 2 * [Ca(OH)₂] = 2 * 2.1 × 10⁻⁴ M ≈ 4.2 × 10⁻⁴ M.
08

Calculate the pH

We can convert the [OH⁻] to pH using the formula: pH = -log[OH⁻] = -log(4.2 × 10⁻⁴) ≈ -3.38. #d.# Calculate [OH⁻] and pH for a solution formed by mixing 20.0 mL of 0.015 M Ba(OH)₂ with 40.0 mL of 8.2 × 10⁻³ M NaOH
09

Calculate the moles of each base

Moles of Ba(OH)₂ = (0.015 M) * (0.020 L) = 3.0 × 10⁻⁴ mol. Moles of NaOH = (8.2 × 10⁻³ M) * (0.040 L) ≈ 3.3 × 10⁻⁴ mol.
10

Calculate the total number of moles of OH⁻

Moles of OH⁻ from Ba(OH)₂ = 2 * moles of Ba(OH)₂ = 2 * (3.0 × 10⁻⁴ mol) = 6.0 × 10⁻⁴ mol. Moles of OH⁻ from NaOH = moles of NaOH = 3.3 × 10⁻⁴ mol. Total moles of OH⁻ = 6.0 × 10⁻⁴ mol + 3.3 × 10⁻⁴ mol = 9.3 × 10⁻⁴ mol.
11

Calculate the total concentration of OH⁻

Total volume = 20.0 mL + 40.0 mL = 60.0 mL = 0.060 L. Total [OH⁻] = (total moles of OH⁻) / (total volume) = (9.3 × 10⁻⁴ mol) / (0.060 L) ≈ 1.55 × 10⁻² M.
12

Calculate the pH

We can convert the [OH⁻] to pH using the formula: pH = -log[OH⁻] = -log(1.55 × 10⁻²) ≈ -1.81.

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Most popular questions from this chapter

Indicate whether each of the following statements is true or false. For each statement that is false, correct the statement to make it true. (a) Acid strength in a series of \(\mathrm{H}-\) A molecules increases with increasing size of \(\mathrm{A} .\) (b) For acids of the same general structure but differing electronegativities of the central atoms, acid strength decreases with increasing electronegativity of the central atom.(c) The strongest acid known is HF because fluorine is the most electronegative element.

Is each of the following statements true or false? (a) All strong acids contain one or more H atoms. (b) A strong acid is a strong electrolyte. (c) A \(1.0-M\) solution of a strong acid will have \(\mathrm{pH}=1.0 .\)

Ephedrine, a central nervous system stimulant, is used in nasal sprays as a decongestant. This compound is a weak organic base: $$\mathrm{C}_{10} \mathrm{H}_{15} \mathrm{ON}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{C}_{10} \mathrm{H}_{15} \mathrm{ONH}^{+}(a q)+\mathrm{OH}^{-}(a q)$$ A 0.035\(M\) solution of ephedrine has a pH of 11.33 . (a) What are the equilibrium concentrations of \(\mathrm{C}_{10} \mathrm{H}_{15} \mathrm{ON}, \mathrm{C}_{10} \mathrm{H}_{15} \mathrm{ONH}^{+},\) and \(\mathrm{OH}^{-} ?\) (b) Calculate \(K_{b}\) for ephedrine.

The volume of an adult's stomach ranges from about 50 mL when empty to 1 when full. If the stomach volume is 400 mL and its contents have a pH of \(2,\) how many moles of \(\mathrm{H}^{+}\) does the stomach contain? Assuming that all the \(\mathrm{H}^{+}\) comes from \(\mathrm{HCl}\) , how many grams of sodium hydrogen carbonate will totally neutralize the stomach acid?

Write the chemical equation and the \(K_{b}\) expression for the reaction of each of the following bases with water: (a) propylamine, \(\mathrm{C}_{3} \mathrm{H}_{7} \mathrm{NH}_{2} ;\) (b) monohydrogen phosphate ion, \(\mathrm{HPO}_{4}^{2-} ;(\mathbf{c})\) benzoate ion, \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2}^{-}.\)
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