Falling Behind in Studies? Join Our Performance Improvement Batch. Enroll For Free
One of our academic counsellors will contact you within 1 working day.
Click to Chat
1800-1023-196
+91-120-4616500
CART 0
Use Coupon: CART20 and get 20% off on all online Study Material
Welcome User
OR
LOGIN
Complete Your Registration (Step 2 of 2 )
Calculate the pH of a solution obtained by mixing 0.1 litre of pH = 4 and 0.2 litre of pH = 10
ph of strong acid = 4 [H+] = 10-4 = molarity of acid in 0.1L , moles of acid = 0.1* 10-4 ph of base =10 [H+]=10-10 [OH-]=10-14/10-10=10-4 = molarity of base in 0.2L , moles of acid = 0.2*10-4 now nutralization will occur and acid is completely nutralised by base , at the end base remaining due to higher concentration.... now final molariry = remaining moles/total vol = (0.2*10-4-0.1*10-4 ) /0.3 concentration of base =10-4/3 [H+] = 3*10-14/10-4 =3*10-10 ph=-loh[H+] = 10-log3= 9.523
ph of strong acid = 4
[H+] = 10-4 = molarity of acid
in 0.1L , moles of acid = 0.1* 10-4
ph of base =10
[H+]=10-10
[OH-]=10-14/10-10=10-4 = molarity of base
in 0.2L , moles of acid = 0.2*10-4
now nutralization will occur and acid is completely nutralised by base , at the end base remaining due to higher concentration....
now final molariry = remaining moles/total vol = (0.2*10-4-0.1*10-4 ) /0.3
concentration of base =10-4/3
[H+] = 3*10-14/10-4 =3*10-10
ph=-loh[H+] = 10-log3= 9.523
Post Question
Dear , Preparing for entrance exams? Register yourself for the free demo class from askiitians.
points won -