The pKa value of is 5. Calculate the pH of the buffer solution, 1 L of which contains 0.01 M and 0.1 — Ionic Equilibrium Chemistry Question

💡 Solution & Explanation

I need to find the pH of a buffer solution containing a weak base ($NH_4OH$) and its conjugate acid ($NH_4Cl$). **Given Information:** - pKa of $NH_3$ = 5, so Ka = $10^{-5}$ - $[NH_4OH]$ = 0.10 M (weak base) - $[NH_4Cl]$ = 0.01 M (conjugate acid, $NH_4^+$) **Step 1: Find pKb of $NH_3$** Using the relationship: $pK_a + pK_b = 14$ $$pK_b = 14 - 5 = 9$$ So $K_b = 10^{-9}$ **Step 2: Apply Henderson-Hasselbalch Equation for Basic Buffer** For a buffer with a weak base and its conjugate acid: $$pOH = pK_b + \log\frac{[NH_4^+]}{[NH_4OH]}$$ $$pOH = 9 + \log\frac{0.01}{0.10}$$ $$pOH = 9 + \log(0.1)$$ $$pOH = 9 + (-1) = 8$$ **Step 3: Calculate pH** $$pH = 14 - pOH = 14 - 8 = 6$$ **Answer: pH = 6** (Option D) The buffer resists changes in pH due to the equilibrium between the weak base $NH_4OH$ and its conjugate acid $NH_4^+$. The pH of 6 is slightly acidic because the conjugate acid ($NH_4^+$) concentration is much lower than the base concentration, but still present enough to shift pH below 7.