Question

25.0 mL of the acid was used during each titration trial. Fill in the blanks of the table:

Table 2. Titration of unknown acid # 2 (H2SO4) with NaOH Concentration of sodium hydroxide (M): 1,0 M Theoretical [H;SO4)0.44 m Trial #3 7.3 Trial #1 9.3 Trial # 2 9.4 Initial buret volume 50.0mL 40.7 mL 31.3 24.0ml 40.7 mL 31.3 mL Final buret volume Volume of base (mL) (O mL 10 mL 10 mi Volume of base (L) 0.010 mu 0.010 mL O.010 mL Moles of base Moles of acid Acid Concentration (M) Average Concentration (M) Standard Deviation Percent Error NOTES/OBSERVATIONS/CALCULATIONS: Re oe 4 Lab Exercise 9 CHM101 Geiger F19

Answer 1

Tria Conc. of Naole = 1M . Vokeme NaoH = 5 final Bureth- Initial - 50ml - 40.7 = 9.3mL (NOTE: You have calculated Woong Voleure of Base kindly Rectify eroon) Volume Base (L) = 0.0093 L moles Base = molarity x volume = Imol x0.00934 = 0.0093 mol " Pasou +2Nach & Nazsoy + 2120 " Imol base react with long and . 0.6093 mol be 0.0093 mol 4.65 Xlo 3 mal and moes of aid = 4:65x15 3 mol. Acid lonce = mides = 4.65 x103 pero 3 Volunt 25 LX15 = 0.186M Triat2 Volume Nach 40.7 -31.3 mL = 9.4 mL Volume Base (L) = 0.Gogu L moles Bax = 1 mol x0.00944 = 0.0094 mol moles and = 0.0094 = 4.7x10 3 mo 2 Aid conc. = 4.7463 mol = 6.188 M 25xl-3L Trial-3 Volume NooH = 31-3-24 mL = 7.3 mL Volume Base -0.0073L moles Bax - 1 mol x 0.00731

3.65 x 103 mol = 0.0073 mol Moles Acid = 0.0073 mol = 2- And Conc. = 3.65 xlo mol 25x10 L = 0.146M Average come = Trial 1 + 2 +3 = 0.186 t o.188 +0.146 Loric. aug. = 0.1733M 1 - 2 Concou dił (2-x)? Tral CM) 1. 0.186 0.0127 1.6129x16" 2. 0.188 0.0147 2.16098104 3. 0.146 -0.0273 7.4529x104 Sum: 11.2267x104 Standard deviation = (so) I I EXin) N-1 Nino of trials SD = 11.2267 No 4 SD = 0.0237] Ng Lone = 0.1733M Theoretical con = 0.4UM al error= Theoretical-aug. xlo Theoretical (0.44-04433) xlo 0.44 ah error = 60.61.17 of error was found to be 60.61.4.