The pH of a solution obtained by mixing 100 cm³ of 0.1 M HCl solution with 100 cm³ of 0.2 M NaOH solution is

Question

ScanSolve AI Tutor · Accepted Answer

Step 1: Moles of HCl Moles = concentration × volume; Moles of HCl = 0.1~mol/dm³ × 0.1~dm³ = 0.01~mol Step 2: Moles of NaOH Moles = concentration × volume; Moles of NaOH = 0.2~mol/dm³ × 0.1~dm³ = 0.02~mol Step 3: Excess OH⁻ NaOH is in excess, so excess moles of OH⁻ = moles NaOH - moles HCl; 0.02~mol − 0.01~mol = 0.01~mol Step 4: Total volume of solution 100~cm³ + 100~cm³ = 200~cm³ = 0.2~dm³ Step 5: Concentration of OH⁻ [OH^−] = 0.01~mol0.2~dm³ = 0.05~mol/dm³ Step 6: pOH pOH = −log₁₀ (0.05); 0.05 = 5 × 10⁻²; log₁₀ (5 × 10⁻²) = log₁₀ 5 + log₁₀ (10⁻²) = 0.6990 − 2 = −1.3010; pOH = −(−1.3010) = 1.3010 ≈ 1.3 Step 7: pH At 25°C, pH + pOH = 14; pH = 14 − 1.3 = 12.7 Final answer: 12.7

The pH of a solution obtained by mixing 100 cm³ of 0.1 M HCl solution with 100 cm³ of 0.2 M NaOH solution is

Answer

Step-by-step solution

Moles of HCl

Moles of NaOH

Excess OH⁻

Total volume of solution

Concentration of OH⁻

pOH

pH

Try one yourself

Still stuck on this one?

More Chemistry Questions