Question

2. A 0.2349 g sample of an unknown acid requires 33.66 mL of 0.1086 M NaOH for neutralization to a phenolphthalein end point. There are 0.42 mL of 0.1049 M HCI used for back-titration. a. How many moles of OH- are used? moles OH moles H b. How many moles of H+ from HCI? moles H c. How many moles of H+ are there ins solid acid? (Eq. 5) g/mol d. What is the molar mass of the unknown acid? (Eq. 4)

Answer 1

volume of Naoy = 33.66 m2 = 33.66/1000 = 0.03366L volume of HCl = 0.42 mL = 0.42/1000 = 0.0004 L @moles of an = moles Brook. moles fou molarity & volume CL) = 0.1086 m X 0.03366 L - O. 00365 Moles! = 3.65*163 moves 6 mokes of 4 t = moles of her roles of 4t = molgsity & win me = 0.1049 mxo.oooh L = 0.0000419 moles = 4.19 xlos moles ! @ 0.0000 419 moles ghel required to neutralize excess C back tisation of Naoh. These fose, moles q Haoh zege reacted with unknoun acid an any 0.00365 mole - 0.0000419 = 0.003608 mole Reaction Hel & NooH nact tho Imole mole M283 Mole 0.2349 g 0.003608 more moloy mass 9 Unknow = 65. 105 / more Molar Mass - 65.105 g/m