Question

5. Calculate the pH of 250. mL of a 0.0205 M barium hydroxide solution at 25°C. 6. What mass of NaOH should be added to 745 mL of water at 25°C to produce a solution with pH = 9.86? 7. 100.0 mL of a strong acid solution has pH = 3.665. What is the pH of the solution after 75.0 mL of water is added?
Show All Your Work NEATLY!! With EXPLAINATIONS! CARE MORE ABOUT YOUR WORK THAN THE ANSWERS! Answers are 12.61, 2.16, and 3.91 respectively!

Answer 1

Answer to the following questions & » 250 mi of 0.0205 (M) Ba(OH)2 Solution As Ba(OH)2 is a strong base, it is fully dissociated to Bat2 & 8M ions in aqueous sol! Ba(OH)2 Bat2 + 28m . Concentration of Toh) = 2X 0.0205 (M) 0.841 (M) ( As After dissociation, a on are produced So, boh - log [ 5H] = -log (0.041) - 1.387 so pH = 14 - POH = 14-1.387 = 12-613 64 of 250 ml 0.0205(M) Ba(OH), sointi is 12.613 . - POH 12 Now Given pH = 9.86, po H = 14-9.86=4.14; [ ] = 10 - 4.14 Now [NaOH(aa) watcas) + ⓇH Con =7-24x108 (M) 70-5 1 mol OH ion & 1 mol Nash .. 7.24 x 105 mos otion = 7-24 x 105 mol Nooh Now, 7.24 x 105 (M) %otion = 7-24 X505 (M) Na oh . 1000 me admons Naoh soll amount of wash Present is = 7.248 105 x 40 gm . ("Molecular weight of Nam In 745 is 40gm. mot) 40 me aqueous sot, amount of NaoH requires - 7.24 x 10 x 40 x7452,167203 22.1578 10 am 1000 NaoH ~ 2.16 mg Naoh

3 Giren pH = 3.665. () = 103.665. 2.162 x10 4 (M1) milimoles in 100 me 2.162 x 104 (M) soluti = 100 x 2.162 x 154 in 0.0216 milimole. After having added 75 mi to the new volume of the solution will be = (100+75) me - 175 me. " new CH = 0.8216 (M) - 1.234 x 104 CM) 2. pH = -log (+2348104) = 3.91 175