Question

H2O NaCl Balance the equation for the reaction that occurs in this titration: HCI + NaOH - ROUGH TITRATION (with 1 mL increments to reach approximate end point) Rough Titration data 0.00 m Initial volume of NAOH (mL); read the buret Final volume of NaOH (m L); read the buret Volume of NAOH used (mL); subtract 13.00 mL 13 00 mL PRECISE TITRATIONS (with drop-by-drop increments to reach precise endpoint) Trial 3 Trial 2 Trial 1 data data data 13.00 mL Initial volume of NaOH (mL); read the buret 27-90 mL Final volume of NaOH (mL); ready the buret 14.90 mL Volume of NaOH used (mL); (subtract) Trial 3 Trial 2 Trial 1 Calculations Calculated Calculated Values For trial 1, show all calculations. Values Box your answer. Include units. Moles of NaOH used Hint: Start with the mL of NaOH you used. Do a conversion considering you used 0.80 M NAOH Moles of HCI in beaker Hint: Start with moles of NaOH used. Do a conversion using the balanced equation. Liters of HCI in beaker Hint: Do a conversion considering you used 10.0mL of HCI. Remember, 1000 mL = 1 L. Calculated Molarity of unknown HCI solution (M) Hint: Remember that the definition of molarity is moles per liter. So M = moles/L Average molarity of unknown HCI sol'n (M):

Answer 1

SOMS Page No. Date Qua HLL + NaoH Imole 1mste Imele 1 niole 1 lhriser titration(tuial4)), vslua of Na O1 uard 14.90 mL eles of NaOH sd Cairen) Malaudly ef O80 Naon = O.80= Volunu in(e) 80 14.90 0.01192 mste 119.2 O80x 14.90 Jo00 1000 Naon ued Moles D J19 2malis 1 Mole HCU krquin dor Racteow Na OH Na ot ile nequirt Dd19 2nsls alaucd egualuon 0)19 2 nls Molro of Anper= 0.01192 mmalus kratur taker in brakr Liturs of Hu n was 1000 mL 1 LE -3 10 mL -2 lo lo m lo x lo -3 0 0 L

Meay of pnaunHR delutien Volnasl(L) .01192 TA t D.01