Q. \( \mathrm{p}K_a + \mathrm{p}K_b \).

Q: What is the basic relationship between pKand pKb for conjugate acid‑base pair?

For conjugate acid Hand its base A‑, K\cdot Kb = Kw. Taking negative logs gives pK+ pKb = pKw. At 25 °C typically pKw = 14.00, so pKb = 14.00 - pKa.

Q: How do I convert between Kand pKa?

Use pK= -\log_{10} Ka. Conversely K= 10^{-pKa}. Use the same base for logs and keep significant figures consistent.

Q: How do I convert between Kb and pKb?

Use pKb = -\log_{10} Kb. Conversely Kb = 10^{-pKb}. These conversions mirror the acid case.

Q: Why does pK+ pKb equal 14 only sometimes?

Because pK+ pKb = pKw. The value of pKw depends on temperature and ionic strength. At 25 °C pKw \approx 14.00. At other temperatures use the appropriate pKw.

Q: How is the Henderson‑Hasselbalch equation related to pKa?

The Henderson‑Hasselbalch equation is \mathrm{pH} = pK+ \log_{10} \frac{\mathrm{A}^-}{\mathrm{HA}} . It uses pKto relate pH and the ratio of conjugate base to acid concentrations.

Q: What are common mistakes when using pK+ pKb?

Common errors: using non‑conjugate species, assuming pKw=14 at all temperatures, ignoring ionic strength or activity coefficients, and misapplying Henderson‑Hasselbalch outside its valid concentration range.

Answer

Using \(K_a K_b = K_w\) for a conjugate acid–base pair, taking \(-\log\) of both sides gives \(pK_a + pK_b = pK_w\). At \(25^\circ\mathrm{C}\), \(pK_w = 14\). Therefore:

\[
\mathrm{p}K_a + \mathrm{p}K_b = \mathrm{p}K_w = 14
\]

Detailed Explanation

We begin by recalling the definitions of pKa and pKb. By definition, pKa is the negative base 10 logarithm of the acid dissociation constant Ka, and pKb is the negative base 10 logarithm of the base dissociation constant Kb. In symbols,

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Chemistry FAQs

What is the basic relationship between pKand pKb for conjugate acid‑base pair?

For conjugate acid Hand its base A‑, \(K\cdot Kb = Kw\). Taking negative logs gives \(pK+ pKb = pKw\). At 25 °C typically \(pKw = 14.00\), so \(pKb = 14.00 - pKa\).

How do I convert between Kand pKa?

Use \(pK= -\log_{10} Ka\). Conversely \(K= 10^{-pKa}\). Use the same base for logs and keep significant figures consistent.

How do I convert between Kb and pKb?

Use \(pKb = -\log_{10} Kb\). Conversely \(Kb = 10^{-pKb}\). These conversions mirror the acid case.

Why does pK+ pKb equal 14 only sometimes?

Because \(pK+ pKb = pKw\). The value of \(pKw\) depends on temperature and ionic strength. At 25 °C \(pKw \approx 14.00\). At other temperatures use the appropriate \(pKw\).

How is the Henderson‑Hasselbalch equation related to pKa?

The Henderson‑Hasselbalch equation is \( \mathrm{pH} = pK+ \log_{10} \frac{\mathrm{A}^-}{\mathrm{HA}} \). It uses pKto relate pH and the ratio of conjugate base to acid concentrations.

What happens at the half‑equivalence point in weak acid titration?

At half‑equivalence the concentrations of Hand A‑ are equal, so \(\mathrm{pH} = pKa\). That makes pKexperimentally accessible from titration curve.

How do polyprotic acids affect pKand pKb relations?

Polyprotic acids have multiple dissociation steps with different \(pKa_1, pKa_2,\) etc. Each conjugate pair obeys \(pKa_i + pKb_i = pKw\) for that pair. Use the appropriate pKfor the protonation state of interest.

What are common mistakes when using pK+ pKb?

Common errors: using non‑conjugate species, assuming \(pKw=14\) at all temperatures, ignoring ionic strength or activity coefficients, and misapplying Henderson‑Hasselbalch outside its valid concentration range.

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