First picture is data, second picture is quesrion Data Sheet:Unknown acid #9Mass of the Unknown acid: 20.209PH364 3.26369 3:243:263.283.213:303:35Name:Volume NaOH added0.5mlUnknown acid(letter or number on the unknown bottle)#9 Concentration of NaOH(aq):1.Aml15mi2.0ml2.5ml30ml3.5mitoml4.5ml20ml5.Smi6.0m)6.5ml10ml75ml80ml8.5ml 19.0ml9.5ml100ml10 sml11.0m11.5ml12.0ml12.5 ml13.0 mlB S ml14.0 ml14.5 ml15.0 ml#923.513.543.583.643.723.393.7038673.93$4.0014.064:1347 4.204.28 FAT4.364.434.464.644.835.052.939.91Partner's Names:Titration of a weak Acid: Calculations and Post lab questionsVolume NaOH added15.5ml16.0 ml16.5 ml(70ml175ml(letter or number on unknown bottle)0.0995 MpH10.6310:0311.2511.37Mass of Unknown acid:0.0995Concentration of NaOH(aq):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.020592. 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?14. Recall that at the first equivalence point, moles of acid = moles of base added.Volume of NaOH added at the first equivalence point:14.9ML1.48x10 molMoles of NaOH added at the first equivalence point:Volume of NaOH added at the Second equivalence point (if your unknown is polyprotic) only 1 equivalerilepoinVON me of NaOH aF Ist equivalence = 14.9ml5. Use the mass of the acid, the concentration of NaOH and the volume of NaOH added at the firstequivalence point to calculate the molar mass of the acid: (Show your work clearly and include propermass of Unknown -0.2099 Concentration of NaOH = 00995 Munits.)mass of and= 0.2099molar mass of and=1.48x10m ² 141.21 g/molCalculated Molar mass of the unknown acid=141.21 g/mol6. Label the areas of the curve where a buffer is present. Locate the point where pH=pKa (halfway tothe equivalence point.) Label pKa on the graph. If the acid is polyprotic, label the pKa for eachequivalence point on the graph.Done!!! pka=45 9. Calculate the initial concentration of your acid. Recall that the original volume of the solution was0.100 L[Acid]initialUsing your calculated value of K, and the initial concentration of the acid, calculate the initial pH ofyour acid.Set up the ICE table. You may need to use the quadratic formula.Initial pH from the titration:Calculated initial pH:Was the calculated initial pH higher or lower than initial pH from the curve ? Calculate the % error.10. Look up the actual literature value of the pKa of your unknown in an appropriate source. (CRChandbook or the NIST webbook)Literature value of pka =(List both pka's if your acid is polyprotic)Source of the data;Was the actual value higher or lower than the pKa from your graph ? Calculate the % error.Calculated pKa

“Since you have posted a question with multiple sub-parts, we will solve the first three subparts for you. To get the remaining sub-part solved please repost the complete question and mention the sub-parts to be solved.”. Mass of Unknown Acid = 0.209 g Molar mass of unknown aid = 141.21 g/mol Volume of solution = 0.100 L To determine: Initial concentration of Acid Molarity =massMolar mass×1Volume of solution (L)Molarity =0.209 g141.21 g/mol×10.100 LMolarity =0.0148 M Initial Concentration of Acid = 0.0148 MTo Determine: Initial pH of acid Since there is only one equivalence point, the given acid is mono-protic acid. Let the acid can be represented as HA ICE Table for dissociation of weak acid (HA) can be expressed as: HA(aq) ↔ H+(aq) A-(aq) Initial 0.0148 - - Change -x +x +x Equilibrium 0.0148-x x x Expression for equilibrium constant: ka=H+A-HASince, pKa = 4ka = 10-4 = 0.0001Therefore,0.0001 =xx0.0148-x1.48×10-6 - 1×10-4x=x2x2+ 1×10-4x-1.48×10-6 =0On solving the quadratic equation, we getx=0.00116758 & x =−0.00126758 (neglected as concentration can never be negative)x=H+=0.00116758pH =-log(H+)pH=-log(0.00116758)pH=2.93 Calculate pH = 2.93 Initial pH of the titration curve. Initial pH of the titration would be the pH measured at Volume of NaOH =0.0 mL Since, in the data Initial pH measured is at Volume of NaOH = 0.5 mL, The Initial pH of the titration would be 3.26 Calculated pH = 2.93 So, calculated Initial pH is lower than the Initial pH from the titration curve.% error in pH can be calculated as: %error =Calculated pH -Experimental pHCalculated pH×100%error =2.93 -3.262.93×100%error =11.26 %