pH and pK
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AP Chemistry › pH and pK
For the weak acid HF in water, $K_a=1.0\times10^{-4}$ at $25^\circ\text{C}$. Which relationship between $K_a$ and $pK_a$ for HF is correct?
$pK_a=\log(K_a)$
$pK_a=-\log([\text{H}^+])$
$pK_a=\dfrac{1}{K_a}$
$pK_a=\log\left(\dfrac{1}{K_a}\right)+1$
$pK_a=-\log(K_a)$
Explanation
This question assesses the skill of pH and pK. The pKa is the negative logarithm of the acid dissociation constant Ka, reflecting the strength of the acid where lower pKa indicates stronger acids due to greater dissociation. Similarly, pH is -log[H+], so both are logarithmic measures that allow relative comparisons without full calculations. For HF with $Ka=1.0×10^{-4}$, pKa=4, illustrating how pKa inversely relates to Ka's magnitude. A tempting distractor is choice A, which incorrectly uses pKa=log(Ka) without the negative sign, leading to positive values that don't align with typical pKa ranges for weak acids. Remember as a transferable strategy that lower pKa means stronger acid because it corresponds to a larger Ka.
A weak acid HA has $K_a=1.0\times10^{-5}$ at $25^\circ\text{C}$. Which value is closest to $pK_a$ for HA?
5
$5.0\times10^{-1}$
10
1
$1.0\times10^{-5}$
Explanation
This question assesses the skill of pH and pK. The pKa is calculated as -log(Ka), transforming the exponential Ka into a linear scale for easier comparison of acid strengths. Lower pKa values indicate stronger acids due to the inverse logarithmic relationship. For $Ka=1.0×10^{-5}$, pKa=5, as $-log(10^{-5}$)=5. A tempting distractor is choice D, which is the Ka value itself, confusing pKa with the dissociation constant rather than its logarithmic form. Remember as a transferable strategy that lower pKa means stronger acid because it corresponds to a larger Ka.
Two weak bases, $\text{NH}_3$ and B, are each dissolved separately in water to make solutions of the same initial concentration at $25^\circ\text{C}$. The bases have $K_b$ values $K_b(\text{NH}_3)=1.0\times10^{-5}$ and $K_b(\text{B})=1.0\times10^{-3}$. Which statement correctly compares the $pK_b$ values?
$pK_b(\text{NH}_3) < pK_b(\text{B})$ because $\text{NH}_3$ has the smaller $K_b$.
$pK_b(\text{NH}_3) > pK_b(\text{B})$ because $\text{NH}_3$ has the larger $K_b$.
$pK_b(\text{NH}_3) = pK_b(\text{B})$ because both are weak bases.
$pK_b(\text{NH}_3) < pK_b(\text{B})$ because $\text{NH}_3$ has the larger $K_b$.
$pK_b(\text{NH}_3) > pK_b(\text{B})$ because $\text{NH}_3$ has the smaller $K_b$.
Explanation
This question assesses the skill of pH and pK. The pKb measures base strength logarithmically as -log(Kb), where a smaller pKb indicates a larger Kb and stronger base. For bases of the same concentration, relative pKb values allow comparison without calculating exact pOH or pH. Here, NH3 with smaller Kb has higher pKb than B, indicating it's weaker. A tempting distractor is choice A, which incorrectly states pKb(NH3) < pKb(B) despite NH3's smaller Kb, misunderstanding the inverse logarithmic relationship. Remember as a transferable strategy that lower pKb means stronger base because it corresponds to a larger Kb.
A weak base BOH has $pK_b=4.0$ at $25^\circ\text{C}$. Which statement correctly describes the magnitude of $K_b$?
$K_b=4.0\times10^{-4}$
$K_b=1.0\times10^{-4}$
$K_b=1.0\times10^{4}$
$K_b=1.0\times10^{-10}$
$K_b=4.0\times10^{4}$
Explanation
This question assesses the skill of pH and pK. The pKb is -log(Kb), so Kb can be found as $10^{-pKb}$, linking the logarithmic pKb to the base dissociation constant. Lower pKb indicates stronger bases with larger Kb values. For pKb=4.0, $Kb=10^{-4}$$=1.0×10^{-4}$. A tempting distractor is choice A, which uses $4.0×10^{-4}$ by mistakenly incorporating the digit 4 without proper logarithmic calculation. Remember as a transferable strategy that lower pKb means stronger base because it corresponds to a larger Kb.
A student prepares separate $0.10,\text{M}$ aqueous solutions of two weak acids, HA and HB, at the same temperature. The acids have the following $K_a$ values: $K_a(\text{HA})=1.0\times10^{-4}$ and $K_a(\text{HB})=1.0\times10^{-6}$. Which statement correctly compares the pH values of the two solutions?
The two solutions have the same pH because both are weak acids.
The two solutions have the same pH because both solutions have the same initial concentration.
The HA solution has a lower pH than the HB solution.
The HA solution has a higher pH than the HB solution.
The HB solution has a lower pH than the HA solution because HB has the smaller $K_a$.
Explanation
This question assesses the skill of pH and pK. The pH of a solution measures the concentration of H+ ions on a logarithmic scale, where lower pH indicates higher acidity. For weak acids, pKa is defined as -log(Ka), so a lower pKa corresponds to a larger Ka and thus a stronger acid that dissociates more, producing more H+ and lowering the pH. Comparing two acids of the same concentration, the one with the larger Ka (or lower pKa) will have a lower pH because it generates more H+ ions relative to the weaker acid. A tempting distractor is choice E, which incorrectly states that the HB solution has a lower pH because HB has the smaller Ka, confusing acid strength with the magnitude of Ka. Remember as a transferable strategy that a lower pKa means a stronger acid because it corresponds to a larger Ka, leading to lower pH in solutions of equal concentration.
A weak acid HA is dissolved in water to make an aqueous solution of fixed concentration. The $K_a$ of HA is then changed from $1.0\times10^{-6}$ to $1.0\times10^{-4}$ (all other conditions unchanged). Which statement best describes how the pH of the solution changes?
The pH increases because $pK_a$ increases.
The pH stays the same because $K_a$ does not affect pH.
The pH decreases because the acid is stronger.
The pH stays the same because the concentration is fixed.
The pH increases because the acid is weaker.
Explanation
This question assesses the skill of pH and pK. Increasing Ka means decreasing pKa, indicating a stronger acid that dissociates more, producing higher [H+] and lower pH. The pH is -log[H+], so changes in strength directly affect pH logarithmically. At fixed concentration, larger Ka lowers pH. A tempting distractor is choice A, which incorrectly states pH increases with stronger acid, misunderstanding that stronger acids decrease pH. Remember as a transferable strategy that lower pKa means stronger acid because it corresponds to a larger Ka, leading to lower pH.
A weak base B has $K_b=1.0\times10^{-4}$ at $25^\circ\text{C}$. A student claims, “Because $pK_b=4.0$, the pH of a solution of B must be 4.0.” Which statement best evaluates the claim?
The claim is incorrect because $pK_b$ is a property of the base, whereas pH depends on the extent of reaction and the solution concentration.
The claim is incorrect because $pK_b$ must always be greater than 7.
The claim is correct because $pK_b$ equals $-\log[\text{OH}^-]$ for any solution.
The claim is incorrect because $pK_b$ is defined as $\log(K_b)$, not $-\log(K_b)$.
The claim is correct because $pK_b$ equals the pH of any solution of that base.
Explanation
This question assesses the skill of pH and pK. The pKb is -log(Kb), a fixed property of the base indicating strength, while pH depends on [OH-] which varies with concentration and dissociation extent. Logarithmic relationships allow comparing strengths, but pH isn't equal to pKb. The claim equates pKb directly to pH, ignoring these factors. A tempting distractor is choice A, which incorrectly assumes pKb equals pH for any base solution, confusing the base's property with the solution's property. Remember as a transferable strategy that lower pKb means stronger base because it corresponds to a larger Kb, but pH requires concentration considerations.
A weak base B is dissolved in water. Its base-dissociation constant is $K_b = 1.0\times 10^{-3}$ (so $pK_b = 3.00$). Which statement is correct?
$pK_b$ increases when more base is added because $pK_b$ depends on concentration.
Because $K_b < 1$, B is a strong base and the pH will be close to 14.
A larger $pK_b$ would indicate a stronger base and a higher pH, assuming the same initial concentration.
A smaller $pK_b$ would indicate a stronger base and thus a higher pH, assuming the same initial concentration.
Because $pK_b = 3.00$, the solution must have $\text{pOH} = 3.00$.
Explanation
This problem involves pH and pK relationships for bases. The pKb value is the negative logarithm of the base dissociation constant (Kb), where pKb = -log(Kb). A smaller pKb corresponds to a larger Kb, indicating a stronger base that accepts protons more readily and produces more OH- ions. Since stronger bases create more hydroxide ions at the same concentration, they result in higher pH values (more basic solutions). Option B incorrectly claims that larger pKb means stronger base, when actually larger pKb means smaller Kb and therefore weaker base. Remember the pattern: smaller pKb means stronger base and higher pH at the same concentration.
A weak acid HA has $K_a = 1.0\times 10^{-7}$. Which relationship correctly gives $pK_a$ for HA?
$pK_a = 1/K_a$, so $pK_a = 1.0\times 10^{7}$.
$pK_a = -\log([\text{HA}])$, so $pK_a$ depends on the initial acid concentration.
$pK_a = -\log([\text{H}^+])$, so $pK_a$ equals the pH of the solution.
$pK_a = \log(K_a)$, so $pK_a = -7.00$.
$pK_a = -\log(K_a)$, so $pK_a = 7.00$.
Explanation
This question tests the mathematical relationship between pH and pK values. The pKa is defined as the negative logarithm (base 10) of the acid dissociation constant: pKa = -log(Ka). Given Ka = 1.0 × 10^-7, we apply this formula: pKa = -log(1.0 × 10^-7) = -(-7) = 7.00. This relationship is fundamental to acid-base chemistry and allows conversion between Ka and pKa values. Option B incorrectly omits the negative sign, while options C, D, and E confuse pKa with other quantities like pH or concentration. Remember: pKa = -log(Ka) is the defining relationship, just as pH = -log[H+].
At $25^\circ\text{C}$, two weak bases are compared: Base 1 has $pK_b = 4.00$ and Base 2 has $pK_b = 6.00$. Equal concentrations of each base are dissolved separately in water. Which statement is correct?
Base 1 produces the more basic solution because its $pK_b$ is smaller.
Base 1 is the weaker base because its $pK_b$ is smaller.
Base 2 produces the more basic solution because its $pK_b$ is larger.
Both solutions have the same pH because both are weak bases.
Both solutions must have $\text{pH}=10$ because $pK_b$ values are given.
Explanation
This question tests understanding of pH and pK for comparing base strengths. Base 1 has pKb = 4.00 while Base 2 has pKb = 6.00, meaning Base 1 has the smaller pKb value. Since pKb = -log(Kb), a smaller pKb corresponds to a larger Kb value, indicating Base 1 is the stronger base. At equal concentrations, the stronger base (Base 1) will produce more OH- ions and create a more basic solution with higher pH. Option A incorrectly states that larger pKb produces a more basic solution, reversing the actual relationship. The key insight: smaller pKb means larger Kb, stronger base, and higher pH when comparing bases at the same concentration.