Question

can you help with the balance equation and also the calculations at the bottom of the sheet thank you

Experiment 17 Acid Base Reactions and Titration Report Sheet Name Section Number Provide the balanced equation for standardizing NaOH below. Table L. Standardization of NaOH Trial Trial 2 Trial 3 Initial mass of flask (g) 85.85 92.9285.80 Mass of flask with KHP (g) Initial buret reading of NaOH (mL) 11.87 19.21 4.21 Initial-buret-reading of NaOH (mL) Final buret reading of NaOH (mL) 15.98 33.15 Mass of KHP sample titrated (g) 1.0 Volume of NaOH solution used (mL) Volume of NaOH solution used (L) Moles of KHP used in each sample (mol) Moles of NaOH reacting (mol) Molarity of NaOH solution (M) Average molarity of standardized NaOH (M) This value will be used in calculations for part B of the experiment. Provide one set of calculations from the mass of KHP to the molarity of NaOH for Trial 1 in the space below.

Answer 1

Acid-Base reaction • Reaction: titration: + Naoh - +HO ml) (8 HSO 4 K + Naon - C8H4 04 K Na + H₂O standardization of Naoy - A Initial buret reading of NaOH (hel) = 1.87 ml Final buret reading of Naon (nel) = 15.98 ml * Volume of Naon solution used = Final – initial = 15.98 -1.87 = 14.11 ml À volume of Naoh solution used - 14.11 = 0.01411 L (ia L - 1000 Mass of kohe sample titrated (g) = 1.01 g Moles of kup med in each sample (mol) = m = lot. 1 20422 M= molar mass of Kup= 204.22 glud. :.= m = 10 11 12 = 4,94X103 moles. ☆ Molasity of NaoH from ear. ci); 1 mol of kHP reacts with Imol of Naon in the titration. This in' of Naon = 4.94x18ml Molarity of Naon = n = 4.94X103 = 0.35m. v 0.014112 Similarly trial 2 & trial 3 can be calculated Trial 2 Trial Initial buret reading of Noor 9.21 hu 4.21 mL Final buset reading of Nach 33.13 mL 27.85 ml vol of Naon med (ml) 23.92 mL 23.64 ml vol. of Nach med (1) 0.02392 L 0.02392 L Mass of KIP 1.06 g 1.079 moles of kap 5.19810 nele 5.24x10 Pro 0.22M molarity of Naon 0.35 +0.22+0:22 Average -=0.26M Molarity of Noon - 3 0.22 m.

Balanced equation ss HCl + Naon Nall + H₂O Initial bunet reading of Naou (rul) 1.20 ml Final bunet reading of Naon (mu)= 3140 ml vol. of Naon used (ncl) = 3020nd Vol. of Naon used in (L) = 30.20 = 0.03020 L 1000 slandardized molant, of NaOH (from Table 1) = 0,2619 moles of Naon need = molarity X 0.261600, 7962018 ast moles of hcl reacting, moles of Naon = moles of HCl from egr moles of Hel reaching & 1852x16 moles. Molarity of Hel in solution. moles 7.262010² - 0.31 in. = Volume 0.025 Sinullenly from for trial 233. Tid 2 Triad 3 Initial bust reading of NaOH 1.28 2.49 Final buret reading it Naon 29.98 32.47 vol. af Naot med (m 28.7 29.98 volit Naon used (L) 0.0287 0.02998 moles of Naon med 7.462x10 7.795010 moles of reci reacting 7.462× 10² 7.7958103 nolarity of HCl 0.3019 0.31m. .. Average molarity of HCl = 0.31 +0:30 +0.31 : Average Molenity of Hal = 0.31M Actual molarity of recl = x. Actual calculated 1% enor = 2 . X 100 Actual