Plot the titration curve as pH vs. mL of NaOH added using a computer graphing program. Make sureyou have chosen the correct X and Y axes. Please include gridlines and print out a 8 ½ x 11 plot. Youmust attach the graph to the Lab Report...Locate the equivalence point(s) on the titration curve. It is very unlikely that the equivalence point(s)will be actual data points. You will need to either manually draw a smooth curve connecting your datapoints, or select the option in your graphing program that will draw the best-fit curve for you. Then findthe steepest point(s) on the curve. This is/these are the equivalence point(s). Label the equivalencepoint(s) on the curve. (For polyprotic acids, label all the equivaence points.)3. How many equivalence points does your graph show?4. Recall that at the first equivalence point, moles of acid = moles of base added.Volume of NaOH added at the first equivalence point:Moles of NaOH added at the first equivalence point:Volume of NaOH added at the Second equivalence point (if your unknown is polyprotic) Data Sheet:Unknown acid #9Mass of the Unknown acid:(letter or number on the unknown bottle)Volume NaOH addedpH0.5ml 364 3.26mi3.69 3:243:26TEST2.0ml2.5ml3.0ml3.5mltom?4.5ml5.0mltus's6.0m)6.5ml1.0m)7.5ml8.5ml9.0m9.5m1100ml10.5 ml11.0m11.5MI12.0ml12.5 ml13.0 mlB.5 ml14.0 ml145 ml2 Concentration of NaOH(aq):0.20915.0 ml3:283:283:303:353:443,39TSE est3.543.583.643.723.803.8673.93$ 4.004.0614:13473 4.204.28 T4.354.434.464.644.835.052.939.91Volume NaOH added15.5ml16.0 ml16.5ml17.0m117.5ml0.0995MpH10.6310:0311.2511.3711.504

To draw the titration curve of unknown acid with NaOH as pH vs. mLof NaOH added. Please note that here MS Excel is used to plot the graph.The equivalence point on the above graph can be located will be halfway up the vertical portion of the curve. It is shown in the next step.There is only one steep curve in the graph, hence there is only 1 equivalence point. This means our acid is a monoprotic acid. Volume of NaOH at the equivalence point = 14.9 mL Moles of NaOH at the equivalence point = Molarity × Volume (L)= 0.0995 molL × 14.9 ×10-3 L= 1.48 ×10-3 mol