Question

Question 4 2 pts Time Run Attempt du 33 Minute Zoom sions A 15.0 mL sample of 0.200 M hydrocyanic acid (HCN) is titrated with 0.100 M KOH. What is the pH after 5.00 mL of KOH has been added?pka = 9.31 for HCN. 1131 numents 9.31 8.61 200 • Previous Next

Answer 1

HON+ oH ON + H₂O Initial Emmal) 3 05 o C " 15,0ml x 0.200=3 = mmol 0.100m) x 5.00ml-0.5 -0.5 -0.5 to.5 change (mmol mmol) 2.5 o 0.5 1 Equilibrium cmmol) pH = Pka + log ICNJ Гнс] p" 9.31 + dog los) 1. pH=9.31 + ((-0.3010 – 0.8973)) .pte 9.3) + (-0.6989) : The pH = 8.61 after 5.00 ml of KoH has been added.