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# Calculate the pH of a solution obtained by mixing 0.1L of a strong acid solution of pH=4 and 0.2L of a solution of strong base of pH=10

509 Points
10 years ago

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