Question

Part 1: Standardization of the NaOH Solution Be sure to include the units and correct number of sig, figs. trial Mass of KHPhth Vol Nach Vol NaOH, initial 3.28 1 2 3 0.2319 0.2154 0.2205 Vol NaOH, final 14.99 14.50 14.42 h. 29-3.5 11. 1 10.99 11.07 molecular lonic net ionic Calculate the Molarity of NaOH solution. Show one set of calculations. Present clear and concise calculations. Do all three sets on scrap paper first. Report the molarity of the NaOH solution for each of the 3 trials. 2. Molarity of the NaOH solution Chem 161 Laboratory - Leeward CC - Spring 2020

Answer 1

The reaction of KHP with NaOH can be an I written as NaOH + KHP - Nakp + H₂O no. of moles of NaOH present = no. of moles of KHP used a mass of KHP used - KHP mass Molar mass of KHP 204.2g gmolt "TRIAL 1: mass of KHP = 0.23199 moles of kHP = 0.2319 g = 0.00114 mol 200.22 gmott- moles of NaOH =0.00 llumol, vol. used= 3,28 mL I Molarity = 0.00114 mol = 0.00328 L U 0.00328L50348 molle =10.348 M a 10.35 M). I Molecular equation : du KH Cg My Ou + NaOH (aq) Nak CG Hu y + H₂Oleh KHP caq (aqla NET IONIC Total i at tort Nattk* + е к намч 89 а.. Сан, но NET JONIC :-( common cons cancelled ] HGH OU + Orlag) - CG H Q Cags + Hole)