EnteCalculate the pH at the points in the titration of 40.00 mL of 0.425 M NH3 f
ID: 610742 • Letter: E
Question
EnteCalculate the pH at the points in the titration of 40.00 mL of 0.425 M NH3 for the reaction below. NH3(aq) + HNO3 ? NH4+(aq) + NO3?(aq) For NH3, Kb = 1.8x10-5. Enter your answer with 2 decimal places. Enter scientific notation as 1.23E4. (a) When 11.50 mL of 0.470 M HNO3 have been added. (b) When 65.00 mL of 0.470 M HNO3 have been added. r question here...Explanation / Answer
NH3 + H2O ---> NH4+ + OH- 0.425 -x x x Kb = 1.8 x10^ (-5) = x^2/(0.425-x) x = [OH-] =0.00275 moles of OH- = 0.00275 x40/1000 = 0.00011 11.5 ml 0.47 M HNO3 moles of HNO3 = 11.5 x0.47/1000 = 0.005405 = [H+] net H+ moles = 0.005405-0.00011 = 0.005295 [H+] = 0.005295 x1000/(40+11.5) = 0.1028 pH = 0.988 b) when 65 ml 0.47 M acid is added H+ moles = 65 x0.47/1000 = 0.03055 net H + moles = 0.03055-0.00011 = 0.03044 [H+] = 0.03044 x1000/(65+40) = 0.289 pH = 0.54
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