Question

Can someone please help me fill out this table with appropriate data ?

Acid-Base Titrations Concentration of HCI acid (M): 0.1000 M Trial 3 Trial 4 Trial 5 Volume of HCI used (ml) 10 ml 10 ml 10 ml Moles of HCI (mol) 40.57 ml Initial burette reading (ml) 12.20 ml 17.59 ml 23.00 ml Final burette reading (ml) 46.29 ml 17.59 ml Volume of NaOH(ml) 5.27 ml 5.39 ml 5.41 ml Moles of NaOH (mol) Molarity of NaOH solution Average total molarity of NaOH solution

Answer 1

> Reaction between Hel and NaOH in Hd + NaOH - Nad + H₂O i mol Hd reacts with Imol wash molarity = number of moles volume in litres = number of moles= molarity x volume in litres = molarity & volume in ml x 100

At the equivalence point / at the end of titration number of moles of Hd = number of moles of Nach Trial 3: moles of Hd = molarity of Helx volume in ml x 103 = 0.1000 m x 10 mL x 103 - Comol. (: M=mo!) -3 moles of NaOH = 10 mol moles of NaOH molarity of Naot = volume in mLx 103 L 3 ML lo mo 5.72 x 103 L = 0.175M

ML Trial-4: moles of Hl = 0.1000 m x 10 mL x 10 L = 103 mol. moles of NaOH = 103 mol. 103 mol molarity of Naot solution = 5.39 x 103 L = 0.1855 M. Trial-5: moles of Hel = 0.1000 m x 10 mL x 103 L A ML -3 - 10 mola moles of NaOH = 10² mol. midarity of Nach = 10 mol/S. 41 80 103 L = 0.1848 M.