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?
$pK_a=-\log([\text{H}^+])$
$pK_a=-\log(K_a)$
$pK_a=\log(K_a)$
$pK_a=\log\left(\dfrac{1}{K_a}\right)+1$
$pK_a=\dfrac{1}{K_a}$
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
$1.0\times10^{-5}$
$5.0\times10^{-1}$
1
10
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.
A weak base BOH has $pK_b=4.0$ at $25^\circ\text{C}$. Which statement correctly describes the magnitude of $K_b$?
$K_b=1.0\times10^{-10}$
$K_b=4.0\times10^{4}$
$K_b=1.0\times10^{4}$
$K_b=4.0\times10^{-4}$
$K_b=1.0\times10^{-4}$
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 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 correct because $pK_b$ equals the pH of any solution of that base.
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 incorrect because $pK_b$ is defined as $\log(K_b)$, not $-\log(K_b)$.
The claim is correct because $pK_b$ equals $-\log[\text{OH}^-]$ for any solution.
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 acid HA has $K_a = 1.0\times 10^{-7}$. Which relationship correctly gives $pK_a$ for HA?
$pK_a = -\log(K_a)$, so $pK_a = 7.00$.
$pK_a = -\log([\text{H}^+])$, so $pK_a$ equals the pH of the solution.
$pK_a = 1/K_a$, so $pK_a = 1.0\times 10^{7}$.
$pK_a = -\log([\text{HA}])$, so $pK_a$ depends on the initial acid concentration.
$pK_a = \log(K_a)$, 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 2 produces the more basic solution because its $pK_b$ is larger.
Both solutions must have $\text{pH}=10$ because $pK_b$ values are given.
Base 1 is the weaker base because its $pK_b$ is smaller.
Base 1 produces the more basic solution because its $pK_b$ is smaller.
Both solutions have the same pH because both are weak bases.
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.
Two separate $0.10,\text{M}$ aqueous solutions are prepared at the same temperature: Solution 1 contains weak acid HX with $K_a = 1.0\times 10^{-4}$, and Solution 2 contains weak acid HY with $K_a = 1.0\times 10^{-6}$. Which comparison is correct?
Both solutions have the same pH because both are $0.10,\text{M}$.
Solution 2 has a lower pH because $K_a$ is smaller.
Solution 1 has a higher pH because $K_a$ is larger.
Solution 1 has a lower pH because HX is the stronger acid (larger $K_a$) at the same concentration.
Both solutions have pH $=4$ because $K_a$ values are powers of ten.
Explanation
This question requires comparing pH and pK values for two weak acids. Since Ka represents the acid dissociation constant, a larger Ka value indicates a stronger acid that produces more H+ ions in solution. HX has Ka = 1.0 × 10^-4 while HY has Ka = 1.0 × 10^-6, making HX the stronger acid (its Ka is 100 times larger). At the same initial concentration, the stronger acid HX will produce more H+ ions and therefore have a lower pH than the weaker acid HY. Option A incorrectly states that larger Ka leads to higher pH, when the opposite is true. The key principle: larger Ka means stronger acid and lower pH when comparing acids at the same concentration.
A $0.10,\text{M}$ solution of the weak acid HA has $K_a = 1.0\times 10^{-5}$ (so $pK_a = 5.00$). Which statement best relates the value of $pK_a$ to the acidity of the solution?
$pK_a$ is the negative logarithm of the initial acid concentration, so changing concentration changes $pK_a$.
Because $pK_a = 5.00$, the solution must have $\text{pH} = 5.00$.
A smaller $pK_a$ would indicate a stronger acid and thus a more acidic solution (lower pH), assuming the same initial concentration.
A larger $pK_a$ indicates a stronger acid and thus a lower pH, assuming the same initial concentration.
Because $K_a$ is less than 1, HA is a strong acid and the solution will have a very low pH.
Explanation
This question tests understanding of pH and pK relationships. The pKa value represents the negative logarithm of the acid dissociation constant (Ka), where pKa = -log(Ka). A smaller pKa value corresponds to a larger Ka value, which indicates a stronger acid that dissociates more completely in water. Since stronger acids produce more H+ ions at the same concentration, they result in lower pH values (more acidic solutions). The incorrect option C reverses this relationship by claiming larger pKa means stronger acid, when actually larger pKa means smaller Ka and therefore weaker acid. Remember: lower pKa means stronger acid and lower pH at the same concentration.
A student states: “Because a solution has p$K_a=5.0$, its pH must be 5.0.” The student is referring to a $0.10,\text{M}$ aqueous solution of a weak monoprotic acid HA with p$K_a=5.0$ at $25^\circ\text{C}$. Which statement best evaluates the student’s claim?
The claim is incorrect because p$K_a$ must always be greater than 7 for a weak acid.
The claim is correct because p$K_a$ depends only on the initial concentration of HA.
The claim is incorrect because p$K_a$ is a measure of acid strength, not the actual pH of a particular solution.
The claim is correct because p$K_a$ is defined as the pH of the acid solution.
The claim is correct because p$K_a$ equals $-\log[\text{H}^+]$ for any acid solution.
Explanation
This question addresses a common misconception about pH and pK values. The pKa is a property of the acid itself, representing its strength through the relationship pKa = -log(Ka). The pH of a solution depends on both the acid's strength (Ka or pKa) and its concentration, requiring equilibrium calculations. For a 0.10 M solution of an acid with pKa = 5.0, the actual pH will be less than 5.0 (typically around 3.0) because the acid only partially dissociates. Choice A incorrectly defines pKa as the pH of the solution, confusing an intrinsic acid property with a solution property. Remember: pKa is a constant for a given acid at a specific temperature, while pH varies with concentration.
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 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 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 $\text{NH}_3$ 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.