The pH of a solution obtained by mixing 50 mL of 1 M HCl and 30 mL of 1 M NaOH is x × 10 . The value — Ionic Equilibrium Chemistry Question

💡 Solution & Explanation

**Step 1: Calculate moles of HCl and NaOH** - Moles of HCl = 1 M × 0.050 L = 0.050 mol - Moles of NaOH = 1 M × 0.030 L = 0.030 mol **Step 2: Determine the limiting reagent** HCl + NaOH → NaCl + H₂O NaOH is the limiting reagent. After neutralization: - Moles of excess HCl = 0.050 - 0.030 = 0.020 mol **Step 3: Calculate concentration of excess HCl** - Total volume = 50 + 30 = 80 mL = 0.080 L - [H⁺] = 0.020 mol / 0.080 L = 0.25 M = 2.5 × 10⁻¹ M **Step 4: Calculate pH** pH = -log[H⁺] = -log(2.5 × 10⁻¹) pH = -[log 2.5 + log 10⁻¹] pH = -[0.3979 - 1] pH = -(-0.6021) pH = 0.6021 **Step 5: Express pH in the form x × 10⁻⁴** pH = 0.6021 = 6021 × 10⁻⁴ Therefore, x = 6021.00