The OH concentration in a mixture of 5.0 mL of 0.0504 M NH Cl and 2 mL of 0.0210 M NH solution is X — Ionic Equilibrium Chemistry Question
Question
The OH concentration in a mixture of 5.0 mL of 0.0504 M NH Cl and 2 mL of 0.0210 M NH solution is X × 10 M. The value of x is _______. (Nearest integer) [Given Kw = 1 × 10 and K = 1.8 × 10 ] – 4 3 -6 -14 b -5
💡 Solution & Explanation
**Step 1: Calculate moles of NH₄Cl and NH₃** - Moles of NH₄Cl = 0.0504 M × 5.0 mL = 0.252 mmol - Moles of NH₃ = 0.0210 M × 2 mL = 0.042 mmol **Step 2: Find total volume and concentrations** - Total volume = 5.0 + 2 = 7.0 mL - [NH₄⁺] = 0.252/7.0 = 0.036 M - [NH₃] = 0.042/7.0 = 0.006 M **Step 3: Apply Henderson-Hasselbalch equation** This is a buffer solution. Use: $$K_b = \frac{[NH_4^+][OH^-]}{[NH_3]}$$ $$1.8 × 10^{-5} = \frac{0.036 × [OH^-]}{0.006}$$ **Step 4: Solve for [OH⁻]** $$[OH^-] = \frac{1.8 × 10^{-5} × 0.006}{0.036}$$ $$[OH^-] = \frac{1.8 × 10^{-5} × 1}{6}$$ $$[OH^-] = 0.3 × 10^{-5} = 3 × 10^{-6} \text{ M}$$ **Step 5: Express in form X × 10⁻⁵ M** $$[OH^-] = 3 × 10^{-6} = 0.3 × 10^{-5} \text{ M}$$ Therefore, x = **3.00** (since 3 × 10⁻⁶ = 3.00 × 10⁻⁶ M, or x = 3).