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Chapter 10: Problem 40

Calculate the \(\mathrm{pH}\) of each solution given the following: a. \(\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]=1 \times 10^{-8} \mathrm{M}\) b. \(\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]=5 \times 10^{-6} \mathrm{M}\) c. \(\left[\mathrm{OH}^{-}\right]=1 \times 10^{-2} \mathrm{M}\) d. \(\left[\mathrm{OH}^{-}\right]=8.0 \times 10^{-3} \mathrm{M}\) e. \(\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]=4.7 \times 10^{-2} \mathrm{M}\) f. \(\left[\mathrm{OH}^{-}\right]=3.9 \times 10^{-6} \mathrm{M}\)

Short Answer

Expert verified
a) 8, b) 5.3, c) 12, d) 11.9, e) 1.3, f) 8.6

Step by step solution

01

- Determine the formula to calculate \(\text{pH}\)

The \(\text{pH}\) of a solution is calculated using the formula: \[ \text{pH} = -\text{log} \big( [\text{H}_{3}\text{O}^{+}] \big) \] where \([\text{H}_{3}\text{O}^{+}]\) is the concentration of hydronium ions in moles per liter (M).
02

- Calculate \(\text{pH}\) for part a

Given: \[ \big [\text{H}_{3}\text{O}^{+} \big ] = 1 \times 10^{-8} \text{ M} \] Using the formula: \[ \text{pH} = -\text{log} \big (1 \times 10^{-8} \big ) \] We get: \[ \text{pH} = 8 \]
03

- Calculate \(\text{pH}\) for part b

Given: \[ \big [\text{H}_{3}\text{O}^{+} \big ] = 5 \times 10^{-6} \text{ M} \] Using the formula: \[ \text{pH} = -\text{log} \big (5 \times 10^{-6} \big ) \] We get: \[ \text{pH} \approx 5.3 \]
04

- Determine the formula to calculate \([\text{H}_{3}\text{O}^{+}]\) from \([\text{OH}^{-}]\)

To calculate \([\text{H}_{3}\text{O}^{+}]\) from \([\text{OH}^{-}]\), use the formula: \[ [\text{H}_{3}\text{O}^{+}] = \frac{10^{-14}}{[\text{OH}^{-}]} \] This relationship is based on the ion product of water \(K_w = 10^{-14} \) at 25掳C.
05

- Calculate \(\text{pH}\) for part c

Given: \[ \big [\text{OH}^{-} \big ] = 1 \times 10^{-2} \text{ M} \] First, calculate \[ \big [\text{H}_{3}\text{O}^{+} \big ]: \] \[ \big [\text{H}_{3}\text{O}^{+} \big ] = \frac{10^{-14}}{1 \times 10^{-2}} = 1 \times 10^{-12} \text{ M} \] Now, calculate the \(\text{pH}\): \[ \text{pH} = -\text{log} \big (1 \times 10^{-12} \big ) \] We get: \[ \text{pH} = 12 \]
06

- Calculate \(\text{pH}\) for part d

Given: \[ \big [\text{OH}^{-} \big ] = 8.0 \times 10^{-3} \text{ M} \] First, calculate \[ \big [\text{H}_{3}\text{O}^{+} \big ]: \] \[ \big [\text{H}_{3}\text{O}^{+} \big ] = \frac{10^{-14}}{8.0 \times 10^{-3}} \approx 1.25 \times 10^{-12} \text{ M} \] Now, calculate the \(\text{pH}\): \[ \text{pH} = -\text{log} \big (1.25 \times 10^{-12} \big ) \] We get: \[ \text{pH} \approx 11.9 \]
07

- Calculate \(\text{pH}\) for part e

Given: \[ \big [\text{H}_{3}\text{O}^{+} \big ] = 4.7 \times 10^{-2} \text{ M} \] Using the formula: \[ \text{pH} = -\text{log} \big (4.7 \times 10^{-2} \big ) \] We get: \[ \text{pH} \approx 1.3 \]
08

- Calculate \(\text{pH}\) for part f

Given: \[ \big [\text{OH}^{-} \big ] = 3.9 \times 10^{-6} \text{ M} \] First, calculate \[ \big [\text{H}_{3}\text{O}^{+} \big ]: \] \[ \big [\text{H}_{3}\text{O}^{+} \big ] = \frac{10^{-14}}{3.9 \times 10^{-6}} \approx 2.56 \times 10^{-9} \text{ M} \] Now, calculate the \(\text{pH}\): \[ \text{pH} = -\text{log} \big (2.56 \times 10^{-9} \big ) \] We get: \[ \text{pH} \approx 8.6 \]

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Key Concepts

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

hydronium ion concentration
Hydronium ion concentration \(\big[[H_{3}O^{+}]\big]\) tells us how acidic a solution is. Hydronium ions form when an acid dissolves in water. We measure the amount of \(H_{3}O^{+}\) in moles per liter (M).
When we need to find the pH of a solution, we use the formula: \(\text{pH} = -\text{log} \big ([H_{3}O^{+}] \big)\). This means we take the negative logarithm (base 10) of the hydronium ion concentration.
For example, if \([H_{3}O^{+}] = 1 \times 10^{-8} \text{ M}\), plugging this into the formula looks like this:
\[ \text{pH} = -\text{log} \big (1 \times 10^{-8} \big ) = 8 \]
  • Small [H_{3}O^{+}] means higher pH (less acidic, more basic).
  • Large [H_{3}O^{+}] means lower pH (more acidic).
hydroxide ion concentration
Hydroxide ion concentration \[ \big [OH^{-} \big] \] indicates the basicity of a solution. Hydroxide ions form when a base dissolves in water. Like hydronium ions, we measure \(OH^{-}\) in moles per liter (M).
To find \[ [H_{3}O^{+}] \] from \[ [OH^{-}] \], use the ion product of water formula:
\[ [H_{3}O^{+}] = \frac{10^{-14}}{[OH^{-}]} \]
Given \[ \big[ OH^{-} \big] = 1 \times 10^{-2} \text{ M} \]:
  • Calculate \[ \big[ H_{3}O^{+} \big]= \frac{10^{-14}}{1 \times 10^{-2}} = 1 \times 10^{-12} \text{ M} \]
  • Now, use the pH formula: \[ \text{pH} = -\text{log} \big (1 \times 10^{-12} \big ) = \text{12} \]
Remember these relationships:
  • If \([OH^{-}]\) is high, \([H_{3}O^{+}]\) is low, making the solution more basic (higher pH).
  • If \([OH^{-}]\) is low, \([H_{3}O^{+}]\) is high, making the solution more acidic (lower pH).
ion product of water
The ion product of water \(\big( K_{w}\big)\) is essential in pH calculations. At 25掳C, \ K_{w} = 1 \times 10^{-14} \). This means that in pure water, \ [H_{3}O^{+}] \times [OH^{-}] = 10^{-14} \text{ M}^{2} \.

This constant helps us understand and convert between \[ [H_{3}O^{+}] \] and \[ [OH^{-}] \]. For example, if we know \[ [OH^{-}] \], we can find \[ [H_{3}O^{+}] \] using: \[ [H_{3}O^{+}] = \frac{10^{-14}}{[OH^{-}]} \]

Let's apply this to another example: Given \[ [OH^{-}] = 3.9 \times 10^{-6} \text{ M} \]
  • First, calculate \[ \big [H_{3}O^{+} \big] = \frac{10^{-14}}{3.9 \times 10^{-6}} \approx 2.56 \times 10^{-9} \text{ M} \]
  • Now, use the pH formula: \[ \text{pH} = -\text{log} \big (2.56 \times 10^{-9} \big ) \approx \text{8.6} \]
This balance of ions is key to understanding how solutions change between acidic and basic.
  • When \ [H_{3}O^{+}] \times [OH^{-}] = 10^{-14} \), the solution is neutral.
  • If \[ [OH^{-}] \] increases, \[ [H_{3}O^{+}] \] must decrease to maintain \(K_{w}\big).
  • If \[ [H_{3}O^{+}] \] increases, \[ [OH^{-}] \] must decrease to maintain \(K_{w}\big).

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Most popular questions from this chapter

Calculate the volume, in milliliters, of a 0.215 M KOH solution that will completely neutralize each of the following: (10.7) a. \(2.50 \mathrm{~mL}\) of a \(0.825 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4}\) solution b. \(18.5 \mathrm{~mL}\) of a \(0.560 \mathrm{M} \mathrm{HNO}_{3}\) solution c. \(5.00 \mathrm{~mL}\) of a \(3.18 \mathrm{M} \mathrm{HCl}\) solution

Identify the conjugate acid-base pairs in each of the following equations: (10.2) a. \(\mathrm{HNO}_{2}(a q)+\mathrm{HS}^{-}(a q) \rightleftarrows \mathrm{NO}_{2}^{-}(a q)+\mathrm{H}_{2} \mathrm{~S}(g)\) b. \(\mathrm{HCl}(a q)+\mathrm{OH}^{-}(a q) \longrightarrow \mathrm{Cl}^{-}(a q)+\mathrm{H}_{2} \mathrm{O}(l)\)

Which of the following are at equilibrium? a. The rate of the forward reaction is twice as fast as the rate of the reverse reaction. b. The concentrations of the reactants and the products do not change. c. The rate of the reverse reaction does not change.

Identify the reactant that is a Br枚nsted-Lowry acid and the reactant that is a Bronsted-Lowry base in each of the following: a. \(\mathrm{CO}_{3}{ }^{2-}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftarrows \mathrm{HCO}_{3}^{-}(a q)+\mathrm{OH}^{-}(a q)\) b. \(\mathrm{H}_{2} \mathrm{SO}_{4}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{HSO}_{4}^{-}(a q)+\mathrm{H}_{3} \mathrm{O}^{+}(a q)\)

Which of the following are not at equilibrium? a. The rates of the forward and reverse reactions are equal. b. The rate of the forward reaction does not change. c. The concentrations of reactants and the products are not constant.
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