3 3 33 10 9 4 3 33 10 10 23 To a 200 ml of 0 1 M weak acid

Last updated: 7/2/2023

3 3 33 10 9 4 3 33 10 10 23 To a 200 ml of 0 1 M weak acid HA solution 90 ml of 0 1 M 23 solution of NaOH be added Now what volume of 0 1 M NaOH be added into above solution so that pH of resulting solution be 5 K HA 105 1 2 ml 2 20 ml 3 10 ml 4 15 ml 3 3 33 x 10 9 4 3 33 10 10 HA 0 1M 200 ml NaOH 90 ml faca fara M 1 fan 0 1 M Nafa facra ap afruit facer a pH 5 K HA 10 1 2 ml 2 20 ml 3 10 ml 4 15 ml