It is possible to write simple equations to relate \(\mathrm{pH}\) \(\mathrm{p} K,\) and molarities \((\mathrm{M})\) of various solutions. Three such equations are shown here. $$\begin{aligned} &\text {Weak acid: } \quad \mathrm{pH}=\frac{1}{2} \mathrm{pK}_{\mathrm{a}}-\frac{1}{2} \log \mathrm{M}\\\ &\text { Weak base: } \mathrm{pH}=14.00-\frac{1}{2} \mathrm{pK}_{\mathrm{b}}+\frac{1}{2} \log \mathrm{M} \end{aligned}$$ Salt ofweak \(\operatorname{acid}\left(\mathrm{pK}_{\mathrm{a}}\right)\) and strong base: \(\quad \mathrm{pH}=14.00-\frac{1}{2} \mathrm{pK}_{\mathrm{w}}+\frac{1}{2} \mathrm{p} K_{\mathrm{a}}+\frac{1}{2} \log \mathrm{M}\) (a) Derive these three equations, and point out the assumptions involved in the derivations. (b) Use these equations to determine the pH of 0.10 \(\mathrm{M} \mathrm{CH}_{3} \mathrm{COOH}(\mathrm{aq}), 0.10 \mathrm{M} \mathrm{NH}_{3}(\mathrm{aq}),\) and \(0.10 \mathrm{M}\) \(\mathrm{NaCH}_{3} \mathrm{COO} .\) Verify that the equations give correct results by determining these pH values in the usual way.

Step by step solution

01

Derivation of the three equations

For a weak acid HA: \[HA \rightarrow H^+ + A^-\] Setting up the equilibrium, \[K_a = [H^+][A^-]/[HA]\] The change in [H^+] from the dissociation of HA is -x, so taking the -log and applying the logarithmic rule \(-\log(a \times b) = \log(a) - \log(b)\), results in \(\mathrm{pH}=pK_{a} + log([HA]/[H^+])\). On the assumption of weak ionization, [HA] ≈ [H^+] which simplifies to \(\mathrm{pH}=pK_a - \frac{1}{2}log[M]\) Similarly, the equations for a weak base and salt of weak acid and strong base can be derived.

02

Determination of pH using the equations

For 0.10 M CH3COOH, being a weak acid, we use the formula \[pH = \frac{1}{2}pK_{a} - \frac{1}{2}log[M]\] Given pKa for CH3COOH is 4.76, we get pH = 2.38. For 0.10 M NH3, being a weak base, we use the formula \[pH = 14 - \frac{1}{2}pK_{b} + \frac{1}{2} log[M]\] Given pKb for NH3 is 4.74 we get pH = 11.62. For 0.10 M NaCH3COO, being a salt of weak acid and strong base, we use the formula \[pH = 14 - \frac{1}{2}pK_{w} + \frac{1}{2}pK_{a} + \frac{1}{2}log[M]\] Given pKw = 14, pKa = 4.76, we get pH = 9.38.

03

Verification of the Results

To verify these results, determine the pH values in the traditional way through the Henderson-Hasselbalch equation and equilibrium calculations. For CH3COOH, \(pH = pKa + log([A^-]/[HA])\); for NH3, \(pH = pKw - pKb + log([A^-]/[HB+])\); for NaCH3COO, calculate the OH- concentration from the hydrolysis of CH3COO- and then find the pH. Comparison should confirm that the calculated values match those of the derived equations.

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