Please wait...
1800-212-7858 (Toll Free)
9:00am - 8:00pm IST all days

or

Thanks, You will receive a call shortly.
Customer Support

You are very important to us

For any content/service related issues please contact on this toll free number

022-62211530

Mon to Sat - 11 AM to 8 PM

200 mL of decimolar H2SO4 is mixed with 200 ml of semimolar NaOH, find the total MV.

Asked by tejas pandey 4th August 2012, 10:16 AM
Answered by Expert
Answer:

The formula use will be:  MH2SO4VH2SO4 + MNaOHVNaOH = MfinalVfinal

 The no. of moles of 0.1 M sulphuric acid or MH2SO4VH2SO4 = 0.1 M x 0.2 l = 0.02 moles

 The no. of moles of 0.5 M NaOH or MNaOHVNaOH = 0.5 M x 0.2 l = 0.1 moles

 So, the total no. of moles or one can say MfinalVfinal = 0.02 + 0.1 = 0.12 moles

 The final volume = 0.2 litres + 0.2 litres = 0.4 litres

 So, the molarity of mixture obtained after mixing = total no. of moles/total volume

           

                                                                         = 0.12/0.4

                                                                     

                                                                         = 0.3 M

 

 

Answered by Expert 30th August 2012, 12:54 PM
Rate this answer
  • 1
  • 2
  • 3
  • 4
  • 5
  • 6
  • 7
  • 8
  • 9
  • 10

You have rated this answer /10

Your answer has been posted successfully!

Chat with us on WhatsApp