Weak Acid and Base Equilibria - AP Chemistry
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What is the conjugate base of acetic acid, CH₃COOH?
What is the conjugate base of acetic acid, CH₃COOH?
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CH₃COO⁻. Loses a proton to form the conjugate base.
CH₃COO⁻. Loses a proton to form the conjugate base.
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How do you find $[OH^-]$ from $pOH$?
How do you find $[OH^-]$ from $pOH$?
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$[OH^-] = 10^{-pOH}$. Inverse logarithm converts pOH back to concentration.
$[OH^-] = 10^{-pOH}$. Inverse logarithm converts pOH back to concentration.
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How is $pK_b$ calculated from $K_b$?
How is $pK_b$ calculated from $K_b$?
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$pK_b = -\text{log}(K_b)$. Negative logarithm converts base constant to pK scale.
$pK_b = -\text{log}(K_b)$. Negative logarithm converts base constant to pK scale.
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Which is stronger: an acid with $K_a = 1.0 \times 10^{-4}$ or $K_a = 1.0 \times 10^{-6}$?
Which is stronger: an acid with $K_a = 1.0 \times 10^{-4}$ or $K_a = 1.0 \times 10^{-6}$?
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The acid with $K_a = 1.0 \times 10^{-4}$ is stronger. Larger $K_a$ value indicates greater acid dissociation.
The acid with $K_a = 1.0 \times 10^{-4}$ is stronger. Larger $K_a$ value indicates greater acid dissociation.
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Identify the effect of dilution on the $pH$ of a weak acid solution.
Identify the effect of dilution on the $pH$ of a weak acid solution.
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Dilution increases $pH$ slightly. Equilibrium shifts right, increasing ionization and pH.
Dilution increases $pH$ slightly. Equilibrium shifts right, increasing ionization and pH.
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What is the $pH$ of a $0.05$ M hydrochloric acid solution?
What is the $pH$ of a $0.05$ M hydrochloric acid solution?
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$pH = 1.3$. Strong acid completely dissociates: $pH = -\log(0.05) = 1.3$.
$pH = 1.3$. Strong acid completely dissociates: $pH = -\log(0.05) = 1.3$.
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Determine the $pH$ of a solution if $[H^+] = 1 \times 10^{-3}$ M.
Determine the $pH$ of a solution if $[H^+] = 1 \times 10^{-3}$ M.
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$pH = 3$. Use $pH = -\log(1 \times 10^{-3}) = 3$.
$pH = 3$. Use $pH = -\log(1 \times 10^{-3}) = 3$.
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What is the relationship between $pH$ and $pOH$ at $25^\text{o}$C?
What is the relationship between $pH$ and $pOH$ at $25^\text{o}$C?
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$pH + pOH = 14$. Sum of pH and pOH equals 14 at standard temperature.
$pH + pOH = 14$. Sum of pH and pOH equals 14 at standard temperature.
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Find $K_a$ given $pK_a = 4.76$.
Find $K_a$ given $pK_a = 4.76$.
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$K_a = 1.74 \times 10^{-5}$. Use $K_a = 10^{-pK_a} = 10^{-4.76} = 1.74 \times 10^{-5}$.
$K_a = 1.74 \times 10^{-5}$. Use $K_a = 10^{-pK_a} = 10^{-4.76} = 1.74 \times 10^{-5}$.
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What is the general expression for the equilibrium constant $K_a$ of a weak acid?
What is the general expression for the equilibrium constant $K_a$ of a weak acid?
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$K_a = \frac{[\text{H}^+][\text{A}^-]}{[\text{HA}]}$. Shows equilibrium between weak acid and its dissociation products.
$K_a = \frac{[\text{H}^+][\text{A}^-]}{[\text{HA}]}$. Shows equilibrium between weak acid and its dissociation products.
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What is the expression for the percent ionization of a weak acid?
What is the expression for the percent ionization of a weak acid?
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Percent ionization = $\frac{[\text{H}^+]}{[\text{HA}]_0} \times 100\text{%}$. Ratio of ionized acid to initial concentration times 100%.
Percent ionization = $\frac{[\text{H}^+]}{[\text{HA}]_0} \times 100\text{%}$. Ratio of ionized acid to initial concentration times 100%.
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What is the Bronsted-Lowry definition of a base?
What is the Bronsted-Lowry definition of a base?
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A base is a proton acceptor. Accepts protons from acids in chemical reactions.
A base is a proton acceptor. Accepts protons from acids in chemical reactions.
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How is $[H^+]$ calculated from $pH$?
How is $[H^+]$ calculated from $pH$?
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$[H^+] = 10^{-pH}$. Inverse logarithm converts pH back to concentration.
$[H^+] = 10^{-pH}$. Inverse logarithm converts pH back to concentration.
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What is the effect of temperature on the value of $K_w$?
What is the effect of temperature on the value of $K_w$?
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$K_w$ increases with temperature. Higher temperature increases water dissociation.
$K_w$ increases with temperature. Higher temperature increases water dissociation.
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What is the $pH$ of a neutral solution at $25^\text{o}$C?
What is the $pH$ of a neutral solution at $25^\text{o}$C?
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$pH = 7$. Equal concentrations of $H^+$ and $OH^-$ at equilibrium.
$pH = 7$. Equal concentrations of $H^+$ and $OH^-$ at equilibrium.
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What is the relationship between $pK_a$ and the strength of an acid?
What is the relationship between $pK_a$ and the strength of an acid?
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Lower $pK_a$ means stronger acid. Lower $pK_a$ corresponds to larger $K_a$ and stronger acid.
Lower $pK_a$ means stronger acid. Lower $pK_a$ corresponds to larger $K_a$ and stronger acid.
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Calculate the $pOH$ of a solution with $[OH^-] = 2 \times 10^{-3}$ M.
Calculate the $pOH$ of a solution with $[OH^-] = 2 \times 10^{-3}$ M.
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$pOH = 2.7$. Use $pOH = -\log(2 \times 10^{-3}) = 2.7$.
$pOH = 2.7$. Use $pOH = -\log(2 \times 10^{-3}) = 2.7$.
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What is the $pH$ of a $0.01$ M sulfuric acid solution assuming complete dissociation?
What is the $pH$ of a $0.01$ M sulfuric acid solution assuming complete dissociation?
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$pH = 2$. Diprotic acid: $[H^+] = 0.02$ M, so $pH = -\log(0.02) = 2$.
$pH = 2$. Diprotic acid: $[H^+] = 0.02$ M, so $pH = -\log(0.02) = 2$.
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Calculate the $pH$ for $[OH^-] = 4 \times 10^{-5}$ M.
Calculate the $pH$ for $[OH^-] = 4 \times 10^{-5}$ M.
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$pH = 9.6$. Find $pOH = 4.4$, then $pH = 14 - 4.4 = 9.6$.
$pH = 9.6$. Find $pOH = 4.4$, then $pH = 14 - 4.4 = 9.6$.
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What is the expression for the equilibrium constant $K_b$ of a weak base?
What is the expression for the equilibrium constant $K_b$ of a weak base?
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$K_b = \frac{[\text{BH}^+][\text{OH}^-]}{[\text{B}]}$. Shows equilibrium between weak base and its protonation products.
$K_b = \frac{[\text{BH}^+][\text{OH}^-]}{[\text{B}]}$. Shows equilibrium between weak base and its protonation products.
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State the formula for calculating $pH$ from $[H^+]$ concentration.
State the formula for calculating $pH$ from $[H^+]$ concentration.
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$pH = -\text{log}([H^+])$. Negative logarithm converts concentration to pH scale.
$pH = -\text{log}([H^+])$. Negative logarithm converts concentration to pH scale.
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How does $pH$ change when a weak acid is diluted?
How does $pH$ change when a weak acid is diluted?
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$pH$ increases slightly. Dilution reduces concentration but increases degree of ionization.
$pH$ increases slightly. Dilution reduces concentration but increases degree of ionization.
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What does a $pH$ greater than $7$ indicate?
What does a $pH$ greater than $7$ indicate?
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The solution is basic. pH above 7 means hydroxide concentration exceeds hydronium.
The solution is basic. pH above 7 means hydroxide concentration exceeds hydronium.
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What does a smaller $K_b$ value indicate about the strength of a base?
What does a smaller $K_b$ value indicate about the strength of a base?
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The base is weaker. Lower $K_b$ means less dissociation and weaker base.
The base is weaker. Lower $K_b$ means less dissociation and weaker base.
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Determine the $pH$ of a $0.1$ M NaOH solution.
Determine the $pH$ of a $0.1$ M NaOH solution.
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$pH = 13$. Strong base: $[OH^-] = 0.1$ M, so $pOH = 1$, $pH = 13$.
$pH = 13$. Strong base: $[OH^-] = 0.1$ M, so $pOH = 1$, $pH = 13$.
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What is the formula for the autoionization of water?
What is the formula for the autoionization of water?
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$2 \text{H}_2\text{O} \rightleftharpoons \text{H}_3\text{O}^+ + \text{OH}^-$. Water molecules exchange protons to form ions.
$2 \text{H}_2\text{O} \rightleftharpoons \text{H}_3\text{O}^+ + \text{OH}^-$. Water molecules exchange protons to form ions.
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What is the value of $K_w$ at $25^\text{o}$C?
What is the value of $K_w$ at $25^\text{o}$C?
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$K_w = 1.0 \times 10^{-14}$. Ion product of water at standard temperature.
$K_w = 1.0 \times 10^{-14}$. Ion product of water at standard temperature.
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What does a larger $K_a$ value indicate about the strength of an acid?
What does a larger $K_a$ value indicate about the strength of an acid?
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The acid is stronger. Higher $K_a$ means greater dissociation and stronger acid.
The acid is stronger. Higher $K_a$ means greater dissociation and stronger acid.
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Determine the $pK_b$ given $K_b = 4.5 \times 10^{-4}$.
Determine the $pK_b$ given $K_b = 4.5 \times 10^{-4}$.
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$pK_b = 3.35$. Use $pK_b = -\log(4.5 \times 10^{-4}) = 3.35$.
$pK_b = 3.35$. Use $pK_b = -\log(4.5 \times 10^{-4}) = 3.35$.
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Find the $pK_a$ given $K_a = 1.8 \times 10^{-5}$.
Find the $pK_a$ given $K_a = 1.8 \times 10^{-5}$.
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$pK_a = 4.74$. Use $pK_a = -\log(K_a) = -\log(1.8 \times 10^{-5}) = 4.74$.
$pK_a = 4.74$. Use $pK_a = -\log(K_a) = -\log(1.8 \times 10^{-5}) = 4.74$.
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