Following solutions were prepared by mixing different volumes of NAOH of HCL different concentrations. (i) 60 mL (M)/(10) HCI+40 mL (M)/(10) NaOH (ii) 55 mL (M)/(10) HCI+45 mL (M)/(10) NaOH (iii) 75 mL (M)/(5) HCI+25 mL (M)/(5) NaOH (iv) 100 mL (M)/(10) HCI+100 mL (M)/(10) NaOH pH of which one of them will be equal to I?

Following solutions were prepared by mixing different volumes of NAOH

1.	Following solutions were prepared by mixing different volumes of NAOH of HCL different concentrations. (i) 60 mL (M)/(10) HCI+40 mL (M)/(10) NaOH (ii) 55 mL (M)/(10) HCI+45 mL (M)/(10) NaOH (iii) 75 mL (M)/(5) HCI+25 mL (M)/(5) NaOH (iv) 100 mL (M)/(10) HCI+100 mL (M)/(10) NaOH pH of which one of them will be equal to I?
Answer» iv (i) (II) (iii) Solution :No of MOLES of HCL `=0.2 xx 75 xx 10^(-3) = 15 xx 10^(-3)` No of mole of `NaOH = 0.2 xx 25 xx 10^(-3) = 5 xx 10^(-3)` No of moles of HCl after mixing `=15 xx 10^(-3) -5 xx 10^(-3)` `:.` Concentration of HCl `= ("No. of moles of HCl")/("Vol in litre") = (10 xx 10^(-3))/(100 xx 10^(-3)) = 0.1M` For (iii) solutiion, pH of 0.1 M HCl = `-log_(10) (0.1) = 1`