Question

Determine the pH during the titration of 50.0 mL of 0.02 M HCl with 0.02 M NaOH after the addition of 35.0 mL and 65.0 mL of titrant.

Answer 1

Number of milimoles of Hel in 50m2 of solution = 50 ml x 0.02M = 0.054*0.02 moit = 0.001 mol = B 1. mmol Numben af milimoles of Naoh in 35 ml. of solutie I mimoles of Naoh in 35 ml af solution = 0.0351X0.02 mall = 0.0007 mol = 0.7 mmol Number of milimoles of Hel left in the solution after addition of 35mL of Naoh base . = 0.3 mmal = (1-0-0-7) mmol 35 mL of Naoh base Total volume of the solution = 50ml+ 35 mL = = 50mL + 35mL = 85 m2 i. 0.3 mmol of Hel are present in 85ml of the solution. Hence, concentration of the Hel in 85 ml of solution is 0.3 mmol - 0.3 mmol _ 0.3x10 mol 85 L 0-085 35x103 nomat = 0.0035 moll ..0.0035 mole of Hel present in the solution often addition of 35 ml af Naot. pH = -log[rig Here, thi] = 0.0035 mole pH = -log [0.0035] =-(-2.45) = 2:45

Therefore, pH of the solution after addition 35 ml of NaOH is 2.45. Now, the pH of the solution after addition of 65.0 m2 of Naoh. Similar to the above : Number of milimoles Hel in 50 ml solution = 1 mmol Number of milimales of Nach in 65 mL of solution = 0.065XX0:02 molt = 0.0013 mol = 1.3 mmol Number of milimoles of Naot left in the aften addition of 65 ml af Naoh solution 65 ml af Naoh = (1.3-1) mmol = 0.3 mmol Totul volume of the solution = 50 mL + 65m2 = 115ml ..0.3 mmol of Nach are present in 115 mL of solution. Hence the concentration of Nach in 115 mL of solution is 0.3 mmol - 0.3 mmol 0.3 X 16 mol 115 ml 0,115 115 X 103 Tanrome = 0.0026 mola ..0.0026 mole af Naot ie ou is present in the solution aften addition of 65 mL of Naoh. TH] [OH) = 1014 1014 [H] = lot ay = 10 0026 = 10-14 2.6x10 3 = 0.384 x 10-11

i pH = -log [HTJ = - dog [0.384x16"]. =-{ log(0.384) + log (16") } =-(-0-415-111 = 11:41 Thus the pH of the solution after addition of 65 mL of Naoh is 11.41