Following solutions were prepared by mixing different volumes of NaOH and HCl of different concentrations : (1) 60 mL (M)/(10) HCl + 40 mL (M)/(10) NaOH 2. 55 mL (M)/(10) HCl + 45 mL (M)/(10) NaOH 75 mL (M)/(5) HCl + 25 mL (M)/(5) NaOH (4) 100 mL (M)/(10) HCl + 100 mL(M)/(10) NaOH pH of which one of them will be equal to 1 ?

Following solutions were prepared by mixing different volumes of NaOH

1.	Following solutions were prepared by mixing different volumes of NaOH and HCl of different concentrations : (1) 60 mL (M)/(10) HCl + 40 mL (M)/(10) NaOH 2. 55 mL (M)/(10) HCl + 45 mL (M)/(10) NaOH 75 mL (M)/(5) HCl + 25 mL (M)/(5) NaOH (4) 100 mL (M)/(10) HCl + 100 mL(M)/(10) NaOH pH of which one of them will be equal to 1 ?
Answer» `(2)` `(1)` `(4)` `(3)` Solution :(1) 4 mL `(M)/(10)` NAOH will neutralize 40 mL of `(M)/(10)` HCl . `(M)/(10)` HCl left unneutralised = 20 mL. TOTAL volume = 100 mL . Dilution = 5 TIMES. In finalsolution, `[HCl]=(M)/(50). pH != 1`. (2) `(M)/(10)`HCl left unneutralised = 10 mL . Total volume = 100 mL . Dilution = 10 times. In final solution, `[HCl]=(M)/(100)=10^(-2)M` `:. pH = 2`. `(3) (M)/(5) ` HCl left unneutralized = 50 mL . Total volume = 100 mL . Dilution = 2 times. In final solution, `[HCl]=(M)/(10)=10^(-1)M`. Hence, pH=1. (4) There is exact NEUTRALISATION. Hence, pH = 7.