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

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