Question:medium

600 mL of 0.04 M HCl is mixed with 400 mL of 0.02 M H2SO4. Find out the pH of resulting solution (Nearest integer).

Updated On: Mar 12, 2026
Show Solution

Correct Answer: 1.4

Solution and Explanation

Step 1: Calculate the moles of HCl.Volume = 600 mL = 0.6 L;
Molarity = 0.04 M;
Moles HCl = Volume × Molarity = 0.6 × 0.04 = 0.024 mol.
Step 2: Calculate the moles  of H2SO4.Volume = 400 mL = 0.4 L;
Molarity = 0.02 M;
Moles H2SO4 = Volume × Molarity = 0.4 × 0.02 = 0.008 mol.
Step 3: Determine total moles of H+ ions.HCl dissociates fully, providing 1 mol of H+ per mol of HCl:
Moles of H+ from HCl = 0.024 mol;
H2SO4 dissociates to provide 2 mol of H+ per mol:
Moles of H+ from H2SO4 = 2 × 0.008 = 0.016 mol;
Total moles of H+ = 0.024 + 0.016 = 0.040 mol.
Step 4: Calculate the total volume of the solution.Total volume = 600 mL + 400 mL = 1000 mL = 1 L.
Step 5: Calculate the final concentration of H+ ions.[H+] = Total moles of H+ / Total volume = 0.040 mol / 1 L = 0.040 M.
Step 6: Calculate the pH of the solution.pH = -log[H+] = -log(0.040) ≈ 1.40.
Verification: The calculated pH ≈ 1.4 falls exactly within the given range (1.4, 1.4).
Was this answer helpful?
0

Top Questions on Equilibrium

Questions Asked in JEE Main exam