Following solutions were prepared by mixing different volume of NaOH and HC1 of different concetrations: (1) 60 mL (M)/(10) HC1 + 40 mL (M)/(10)NaOH (2) 55 mL (M)/(10) HC1 + 45 mL (M)/(10) NaOH (3) 75 mL (M)/(5) HC1 + 25 mL (M)/(5) NaOH 100 mL (M)/(10) HC1 + 100 mL (M)/(10) NaOH pH of which one of them will be equal to 1 ?

Following solutions were prepared by mixing different volume of NaOH a

1.	Following solutions were prepared by mixing different volume of NaOH and HC1 of different concetrations: (1) 60 mL (M)/(10) HC1 + 40 mL (M)/(10)NaOH (2) 55 mL (M)/(10) HC1 + 45 mL (M)/(10) NaOH (3) 75 mL (M)/(5) HC1 + 25 mL (M)/(5) NaOH 100 mL (M)/(10) HC1 + 100 mL (M)/(10) NaOH pH of which one of them will be equal to 1 ?
Answer» 1 2 4 3 Solution :Where `N_(1) V_(1) gt N_(2)V_(2)` Acid is left at the end of REACTION and solutuion will be acidic `N_(FINAL solution) = [H^(+)] = (N_(1)V_(1) - N_(2)V_(2))/(V_(1) + V_(2))` , as basicity and acidity of acid and base is 1 so M = N (1) `[H^(+)] = (6-4)/(100) = (2)/(100)``pH /+ 1` (2) `[H^(+)] = (5.5 - 4.5)/(100) = (1)/(100) = 10^(-2)``pH = 2` (3)`[H^(+)] = (15 - 5)/(100) = (10)/(100) = 10^(-1)``pH = 1` (4) `[H^(+)] = (10 - 10)/(200) = 0`Neutral