Skip to content
Write Brief Answer · Q15

Q.50 mL of 0.05M HNO3HNO_3 is added to 50 mL of 0.025M KOH. Calculate the pH of the resultant solution.

Puducherry TnboardTextbookSubjectiveImportance★★★★★
43% · 30/70 Questions
🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

Step 1. Moles HNO3=50×0.05=2.5HNO_3=50\times0.05=2.5 mmol; moles KOH =50×0.025=1.25=50\times0.025=1.25 mmol.

Step 2. HNO3HNO_3 is in excess: excess =2.5−1.25=1.25=2.5-1.25=1.25 mmol, in a total volume of 50+50=10050+50=100 mL.

Step 3. [H+]=1.25100=0.0125[H^+]=\dfrac{1.25}{100}=0.0125 M =1.25×10−2=1.25\times10^{-2} M. …

Unlock everything free for 14 days

  • Full step-by-step solutions
  • Concept-first explanations
  • Methods, shortcuts & mistakes
  • PYQ mapping + timed mock tests

Full access for 14 days. No credit card required.