InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
Assuming complete dissociation, calculate the `pH` of the following solutions, a. `0.003 M HCl, b. 0.005 M NaOH`, c. `0.002 M HBr, d. 0.002 M KOH` |
|
Answer» Correct Answer - a) `2.52` , b) `11.70` , c) `2.70` , d) `11.30` (i) `0.003 MHCl`: `H_(2)O + HCl harr H_(3)O^(+) + Cl^(-)` Since HCl is completelly ionized `[H_(3)O^(+)] = [HCl]` `rArr [H_(3)O^(+)] = 0.003` Now, `pH = - "log"[H_(3)O^(+)] = -"log" (.003)` `= 2.52` Hence, the pH of the solution is 2.52. (ii) 0.005MNaOH: `NaOH_((aq)) harr Na_((aq))^(+) + HO_((aq))^(-)` `[HO^(-)] = [NaOH]` `rArr [HO^(-)] = .005` `pOH = -"log"[HO^(-)] = -"log"(.005)` `pOH = 2.30` `:. pH = 14-2.30` `= 11.70` Hence, the pH of the solution is `11.70` (ii) `0.002 HBr`: `HBr + H_(1)O harr H_(3)O^(+) + Br^(-)` `[H_(3)O^(+)] = [HBr]` `rArr [H_(3)O^(+)] = .002` `:. pH = -"log" [H_(3)O^(+)]` `= - "log" (0.002)` `= 2.69` Hence the pH of the solution is `2.69`. (iv) `0.002 M KOH`: `KOH_((aq)) harr K_((aq))^(+) OH_((aq))^(-)` `[OH^(-)] = [KOH]` `rArr [OH^(-)] = .002` Now, `pOH = -"log" [OH^(-)]` `= 2.69` `:. pH = 14 - 2.69` `= 11.31` Hence the pH of the solution `11.31` |
|