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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
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