Question

Data Sheet: 1. Concentration of NaOH ALL volume measurements (in mL) must be recorded using two (2) digits to the right of the decimal or they are wrong. (Watch the significant figures.) SHOW ALL OF YOUR CALCULATIONS ON THE NEXT PAGE. 2. Initial volume of acid in buret (left photo) 3. Final volume of acid in buret (right photo) 4. Volume of acid (#3 #2) 5. Volume of acid (#4/1000) 6. Initial NaOH level in buret 7. Final NaOH level in buret 8. Volume NaOH used (#7 - #6) 9. Volume NaOH used (#8/1000) moles 10. Moles of NaOH used (#9 x #1) MoleSbase = Molesacid 11. Moles of Acid Neutralized ( #11 = #10 ) moles 12. Volume of Acid ( #5 ) 13. Molarity of Acid( #11 / #12)

please use the information provided to fill out the first sheet (the data)

Answer 1

1.a. 0.35M

2. 3.4 M

3.a    10^ (-5) M

3.b. pH = - log [10^-5]

= 5$\times$ log 10

= 5
1. Using complete sentences 1. a state the purpose of this experiment. This is an aid-base (strong and against Strong base) volumetric titration using phenolp thalein as indicator to find out the concentration of unknown acid solution. During the process of titration, the pH of the solution varies and as a result end point is reached with a visible colour change (colourless to slight pink). Hence by using the titration formula, - Mix, = M₂ V₂ we can find the unknown concentration. where M, and M₂ as the molarity of acid and base respectively. V and 2 are the volume of acid and base utilized, respectively. This acid-base titration is simply a neutrilization reaction as follows; i H+ + Oh He + NaOH H₂O > H₂O + Nad

job. In order to find out the end points in to acid- base titration, generally pH indicators are used, phenolphthalein is one such pH indicator used in strong acid- strong base titration. On adding phenolphthaleis, we get a noticable colour change from colourless to pink colour, recanatend point or neutrilization point. The pH value at end point depends on the strength of acid and based titrated. The pH sange of phenolptbaleis is 800 to 9.8 If we forgot to add phenolphthalein we are and calculate find can't unable to the end point the unknown concentration. or whicu concentration sur o reg

the lab that State evidence observed from the end point was reached t o be even a od po At end point, the colour of the titration solution changes to pink colour. On adding alkalie to the aid of unknown concentration, the pH of the solution raise and further addition leads to the complete neutrilization of acid. Then a slight addition of a single drop of alkalie result is a rapid increase of pH and the solution undergoes a change in colour at pH prevailing near neutrilization of end point, due to the presence of free hydroxyl zon And solution Alkaline Pink: colourless phenolptbalein

calculations:- 0.25 M Btd NaOH x unknown Hal solution Com phenolpthaleis Final vol of solution Initial Bust Readin Burt Reading CML) 14.1 NaoH 16.) Hel 10 1.9 JM 1109N on. unknown Hell soluli At equivalence point, 1401 mL 0.25 M NaOH = 10mL Mava = mbvb is the titration formulae. Thus the concentration of Ma is Ma Mb Vor at va i nawoor 0.25 x 14 1 10 . = 0.3525M .: conc: of unknown solution of He is 0.35 M

Data sheet 1. conc: of, NaOH : 1 0025 M role 2. Initial vol. of aid is burette : 1.9 mL 3. Final vol. of acid in burette: 11.9 mill! 4. Volume of acid. Sealne 10 mL 5. volume of aud: 0.0 L 6. Initial vol. of NaOH is buset: 2 mL 7. Final vol of NaOH in buset: 16012 8. vol of Naot med : 1401 mL 9. vol of Nach med 0.14 L 10. Mous of Nadt used 0.25 moles. 11. Mous of and Neutsilized 0.35 mous. 12. vot of acid :.010 L 0:35 M 13. molasity of and

Standard Solution : Naol conc: 0.25M vol. of He used (va) = 10mL t lov lot Initial NaOH burete reading = 2 final Naolt burette reading = 16.1 vol. of NaOH used (16) = 16.1-2 =14.1mL

2. . Gin Given that has y Hoola bt Mas Cone of Hel 22.5 m (ma) vol. of Hel = 75mL (va) vol of base = 55 mL (VD) conc: of base = ? (Mb) Using the acid-base tétration formula Mava = Mb VbH Mb= Mava Vb. BH neuon law Jano= 2.5x 750 55 Jers.pl - 3.4 M The concentration of us known baric solution å 3.4 M

z. a Given that conc: of base solution = 0.100M (Mb) conci volume of base solution = 0.100mL (VD) volume of Lake water = 1000mL (2) conc: of Lake water, (molarity) = ? (ML) using the formulae Mb Vb=MLVL ML = Mb Vb VL 0.100x 0.100 1000 = 1x100M " The molasity of acid is Lake water å 1x10BM