Question

Calculate the pH for each of the cases in the titration of 35.0 mL of 0.150 M LiOH(aq) with 0.150 M HCl(aq). Note: Enter your answers with two decimal places. before addition of any HCl: after addition of 13.5 mL HCI: after addition of 20.5 mL HCl: after addition of 35.0 mL HCI: after addition of 45.5 mL HCI: after addition of 50.0 mL HCI:

Answer 1

molarity of Lion (M) = 0.150 m. volume of Lion (V) = 35.0 mL milimples of LiOH (MV = 35.080150 = 5.25 mmol given molarity of HCl = 0.150m Answon a befor addition of Hel POH of Lion solution = -log[DH] = -log(0.150) =-8239) POH = 0.8239 Mence pH = 14 - POM = 14-018239 pH = 13:17 Answon (b) after addition of 12.5 mL M L Then milimoles of HU = 13.5 X 8.150.=20025 mmol aftegn addition of Hel gremaining ammount of Lion Tatal volume concentration = 48.5 ml = (525 -2.095) mmol = 3.295 mmol. the solution = 3510 + 13.5 the solution = 3.225 = of of 0.06649M 48.5 pon of the solution & -log(ort) = -dog (0.06649) = = 1.177 = 14 - 1177 -|-11177) Mence POH pH= 14 - POH [PH = 12.82

(6) After addition of 20.5 mL MW Then milimoles of Hll solution = 20.5 * 0.150 – 3.075 mmol after addition of HCl remaining ammount of Lion = (5125 – 3.075) mmol - 2.175 mmol Tatal volume of solution = 35.0+ 20.5 = 55.5 ml. concentration of the solution = 2.175 = 0.03918 m. = 0 55.5 Por of the solution = -log(on) = -dogl0.03918) = -1-1.4069) pOH = 1.4069 Hence pH = 14-POH = 14-100069 IPH = 4.59 Answorlds after addition of 350 mL of HCI then milimoler of Mel = 3510 x 0.50 = 5125 mmol after addition of Mcl remaining concentration of Lion = 5.95 - 5.95 = 0 Here complete neutralization of the base -acid reaction takes place. Hence TPH = 70007 Answer (e) after addition of 45.5 mL HC milimoles of Ha a 45.5 * 0.150 = 6.825 mmos Hemp ammount (minol) of Mal is more than Lion nence after addition of Hol remaining ammount of Hol is = 6.825 – 5:25 then

Tatal volume concentration of of = 1.575 mmol the solution = the al solution 350 + 45.5 = 80.5 mL = 1.575 - 0.01956M 80.5 = 0.01956 M Mence pH = -log[ht] = - Jog 10.01956) pH =-47080 [ph = 1.717 Answas (f) after addition of 5000 mL ALI milimples of HCl = 500X 0.150 mmol = 7.5 mmpl after addition of Hu snemaiping ammount of HCL = (7.50 -5.25) mmol 5 2.25 Tatal volume of solution = 3500+ 5000 mL = 85 m concentration of remaining Hai = 9:25 = 0.02647 M Honce pH = -log[ht] = -log(0.02697) = -1-1.5772) = 1:577 [pH = 1.56