Titration and Buffers (5A) - MCAT Chemical and Physical Foundations of Biological Systems
Card 1 of 25
State the Henderson–Hasselbalch equation for a weak acid buffer.
State the Henderson–Hasselbalch equation for a weak acid buffer.
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$\text{pH}=\text{p}K_a+\log\left(\frac{[A^-]}{[HA]}\right)$. Derived from the acid dissociation equilibrium, this equation relates pH to the acid's pKa and the ratio of conjugate base to acid concentrations.
$\text{pH}=\text{p}K_a+\log\left(\frac{[A^-]}{[HA]}\right)$. Derived from the acid dissociation equilibrium, this equation relates pH to the acid's pKa and the ratio of conjugate base to acid concentrations.
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What is the relationship between $K_a$ and $K_b$ for a conjugate pair at $25,^{\circ}\text{C}$?
What is the relationship between $K_a$ and $K_b$ for a conjugate pair at $25,^{\circ}\text{C}$?
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$K_aK_b=K_w=1.0\times10^{-14}$. For conjugate acid-base pairs, the product of Ka and Kb equals the water dissociation constant Kw at 25°C.
$K_aK_b=K_w=1.0\times10^{-14}$. For conjugate acid-base pairs, the product of Ka and Kb equals the water dissociation constant Kw at 25°C.
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What is the relationship between $K_b$ and $\text{p}K_b$?
What is the relationship between $K_b$ and $\text{p}K_b$?
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$\text{p}K_b=-\log(K_b)$. pKb measures base strength as the negative logarithm of Kb, with lower pKb indicating stronger bases.
$\text{p}K_b=-\log(K_b)$. pKb measures base strength as the negative logarithm of Kb, with lower pKb indicating stronger bases.
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Find $\frac{[A^-]}{[HA]}$ when $\text{pH}=5.00$ and $\text{p}K_a=4.00$.
Find $\frac{[A^-]}{[HA]}$ when $\text{pH}=5.00$ and $\text{p}K_a=4.00$.
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$\frac{[A^-]}{[HA]}=10$. From rearranged Henderson-Hasselbalch, ratio = 10^(pH - pKa) = 10^(1) = 10.
$\frac{[A^-]}{[HA]}=10$. From rearranged Henderson-Hasselbalch, ratio = 10^(pH - pKa) = 10^(1) = 10.
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What is the relationship between $K_a$ and $\text{p}K_a$?
What is the relationship between $K_a$ and $\text{p}K_a$?
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$\text{p}K_a=-\log(K_a)$. pKa quantifies acid strength as the negative logarithm of Ka, where lower pKa indicates stronger acids.
$\text{p}K_a=-\log(K_a)$. pKa quantifies acid strength as the negative logarithm of Ka, where lower pKa indicates stronger acids.
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Identify the pH relative to $\text{p}K_a$ at the half-equivalence point of a weak acid titration.
Identify the pH relative to $\text{p}K_a$ at the half-equivalence point of a weak acid titration.
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$\text{pH}=\text{p}K_a$. At half-equivalence, [HA] = [A-], so Henderson-Hasselbalch simplifies to pH = pKa.
$\text{pH}=\text{p}K_a$. At half-equivalence, [HA] = [A-], so Henderson-Hasselbalch simplifies to pH = pKa.
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Find the equivalence volume: $25.0,\text{mL}$ of $0.100,\text{M}$ HCl titrated by $0.200,\text{M}$ NaOH.
Find the equivalence volume: $25.0,\text{mL}$ of $0.100,\text{M}$ HCl titrated by $0.200,\text{M}$ NaOH.
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$V_{\text{NaOH}}=12.5,\text{mL}$. Using Ma Va = Mb Vb for 1:1 titration, Vb = (0.100 * 25.0) / 0.200 = 12.5 mL.
$V_{\text{NaOH}}=12.5,\text{mL}$. Using Ma Va = Mb Vb for 1:1 titration, Vb = (0.100 * 25.0) / 0.200 = 12.5 mL.
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Find the pH when $\text{p}K_a=6.30$ and $\frac{[A^-]}{[HA]}=10$.
Find the pH when $\text{p}K_a=6.30$ and $\frac{[A^-]}{[HA]}=10$.
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$\text{pH}=7.30$. Henderson-Hasselbalch gives pH = pKa + log(10) = pKa + 1, yielding 7.30.
$\text{pH}=7.30$. Henderson-Hasselbalch gives pH = pKa + log(10) = pKa + 1, yielding 7.30.
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Find the pH when $\text{p}K_a=4.76$ and $\frac{[A^-]}{[HA]}=1.0$.
Find the pH when $\text{p}K_a=4.76$ and $\frac{[A^-]}{[HA]}=1.0$.
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$\text{pH}=4.76$. Using Henderson-Hasselbalch, when ratio = 1, log(1) = 0, so pH equals pKa directly.
$\text{pH}=4.76$. Using Henderson-Hasselbalch, when ratio = 1, log(1) = 0, so pH equals pKa directly.
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Identify the primary purpose of an indicator in an acid–base titration.
Identify the primary purpose of an indicator in an acid–base titration.
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To signal the endpoint by a visible color change near equivalence. Indicators change color at a specific pH near the equivalence point to visually indicate when to stop the titration.
To signal the endpoint by a visible color change near equivalence. Indicators change color at a specific pH near the equivalence point to visually indicate when to stop the titration.
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Which expression gives the buffer ratio $\frac{[A^-]}{[HA]}$ in terms of pH and $\text{p}K_a$?
Which expression gives the buffer ratio $\frac{[A^-]}{[HA]}$ in terms of pH and $\text{p}K_a$?
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$\frac{[A^-]}{[HA]}=10^{\text{pH}-\text{p}K_a}$. Rearranging the Henderson-Hasselbalch equation isolates the ratio as an exponential function of the pH-pKa difference.
$\frac{[A^-]}{[HA]}=10^{\text{pH}-\text{p}K_a}$. Rearranging the Henderson-Hasselbalch equation isolates the ratio as an exponential function of the pH-pKa difference.
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What is the typical effective buffering range for a weak acid buffer relative to $\text{p}K_a$?
What is the typical effective buffering range for a weak acid buffer relative to $\text{p}K_a$?
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Approximately $\text{p}K_a\pm^1$ pH unit. Buffers are effective within about one pH unit of pKa where the ratio of components allows significant buffering against additions.
Approximately $\text{p}K_a\pm^1$ pH unit. Buffers are effective within about one pH unit of pKa where the ratio of components allows significant buffering against additions.
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At what condition does a buffer have maximum buffering capacity?
At what condition does a buffer have maximum buffering capacity?
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When $\text{pH}=\text{p}K_a$ and $[A^-]=[HA]$. Maximum capacity occurs when the buffer can equally resist acid or base additions, which is at equal concentrations and pH = pKa.
When $\text{pH}=\text{p}K_a$ and $[A^-]=[HA]$. Maximum capacity occurs when the buffer can equally resist acid or base additions, which is at equal concentrations and pH = pKa.
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What is the relationship between $\text{p}K_a$ and $\text{p}K_b$ at $25,^{\circ}\text{C}$?
What is the relationship between $\text{p}K_a$ and $\text{p}K_b$ at $25,^{\circ}\text{C}$?
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$\text{p}K_a+\text{p}K_b=14$. Taking negative logs of the Ka Kb = Kw relation yields pKa + pKb = 14 at 25°C for conjugate pairs.
$\text{p}K_a+\text{p}K_b=14$. Taking negative logs of the Ka Kb = Kw relation yields pKa + pKb = 14 at 25°C for conjugate pairs.
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What is the formula for the equivalence-point volume in a $1{:}1$ acid–base titration?
What is the formula for the equivalence-point volume in a $1{:}1$ acid–base titration?
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$M_aV_a=M_bV_b$. For 1:1 stoichiometry, the product of molarity and volume is equal for acid and base at equivalence.
$M_aV_a=M_bV_b$. For 1:1 stoichiometry, the product of molarity and volume is equal for acid and base at equivalence.
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Identify the dominant species when $\text{pH}<\text{p}K_a$ for a weak acid/conjugate base system.
Identify the dominant species when $\text{pH}<\text{p}K_a$ for a weak acid/conjugate base system.
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The protonated form $HA$ dominates. At pH below pKa, the equilibrium favors the undissociated acid form according to Le Chatelier's principle.
The protonated form $HA$ dominates. At pH below pKa, the equilibrium favors the undissociated acid form according to Le Chatelier's principle.
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Identify the dominant species when $\text{pH}>\text{p}K_a$ for a weak acid/conjugate base system.
Identify the dominant species when $\text{pH}>\text{p}K_a$ for a weak acid/conjugate base system.
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The deprotonated form $A^-$ dominates. At pH above pKa, the equilibrium shifts to favor the dissociated conjugate base form.
The deprotonated form $A^-$ dominates. At pH above pKa, the equilibrium shifts to favor the dissociated conjugate base form.
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What is the pH at the half-equivalence point in a weak acid–strong base titration?
What is the pH at the half-equivalence point in a weak acid–strong base titration?
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$\text{pH}=\text{p}K_a$. At half-equivalence, half the acid is neutralized, making [HA] = [A-], so pH = pKa from Henderson-Hasselbalch.
$\text{pH}=\text{p}K_a$. At half-equivalence, half the acid is neutralized, making [HA] = [A-], so pH = pKa from Henderson-Hasselbalch.
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What is the pH at the equivalence point in a strong acid–strong base titration at $25,^{\circ}\text{C}$?
What is the pH at the equivalence point in a strong acid–strong base titration at $25,^{\circ}\text{C}$?
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$\text{pH}=7$. Neutralization of strong acid by strong base produces neutral salt solution with pH = 7 at 25°C.
$\text{pH}=7$. Neutralization of strong acid by strong base produces neutral salt solution with pH = 7 at 25°C.
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In a weak acid–strong base titration, is the pH at equivalence point less than, equal to, or greater than $7$?
In a weak acid–strong base titration, is the pH at equivalence point less than, equal to, or greater than $7$?
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Greater than $7$. The equivalence point forms the conjugate base of the weak acid, which hydrolyzes to produce a basic solution with pH > 7.
Greater than $7$. The equivalence point forms the conjugate base of the weak acid, which hydrolyzes to produce a basic solution with pH > 7.
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In a weak base–strong acid titration, is the pH at equivalence point less than, equal to, or greater than $7$?
In a weak base–strong acid titration, is the pH at equivalence point less than, equal to, or greater than $7$?
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Less than $7$. The equivalence point forms the conjugate acid of the weak base, which hydrolyzes to produce an acidic solution with pH < 7.
Less than $7$. The equivalence point forms the conjugate acid of the weak base, which hydrolyzes to produce an acidic solution with pH < 7.
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State the equivalence condition for any titration in terms of moles of titrant and analyte.
State the equivalence condition for any titration in terms of moles of titrant and analyte.
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Moles titrant added = stoichiometric moles required for analyte. Equivalence occurs when the titrant provides exactly the moles needed to react completely with the analyte per the stoichiometry.
Moles titrant added = stoichiometric moles required for analyte. Equivalence occurs when the titrant provides exactly the moles needed to react completely with the analyte per the stoichiometry.
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State the formula used to find moles from molarity and volume (with volume in liters).
State the formula used to find moles from molarity and volume (with volume in liters).
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$n=MV$. Molarity times volume in liters gives moles, essential for calculating stoichiometry in titrations.
$n=MV$. Molarity times volume in liters gives moles, essential for calculating stoichiometry in titrations.
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What is the definition of a buffer in aqueous acid–base chemistry?
What is the definition of a buffer in aqueous acid–base chemistry?
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A solution that resists pH change upon small acid/base addition. Buffers maintain stable pH by neutralizing small amounts of added acid or base through reversible reactions involving weak acids/bases and their conjugates.
A solution that resists pH change upon small acid/base addition. Buffers maintain stable pH by neutralizing small amounts of added acid or base through reversible reactions involving weak acids/bases and their conjugates.
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What two components must be present in an effective buffer solution?
What two components must be present in an effective buffer solution?
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A weak acid and its conjugate base (or weak base and conjugate acid). Effective buffers require both components to shift equilibrium and absorb added H+ or OH- ions without significant pH change.
A weak acid and its conjugate base (or weak base and conjugate acid). Effective buffers require both components to shift equilibrium and absorb added H+ or OH- ions without significant pH change.
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