Question

Report Sheet mass of bocx - 0 Part A: Titration of Room Temperature Borax: K Molarity of Standard HCl solution: 0.2 M 3.005g Temperature of room temperature borate mixture (T.): T, ind0 °C T, in 293 K Trial 3 (if done) Trial Trial 2 Final buret reading HCl, mL 18.05 my 32.70ml Initial buret reading HCl, mL 5.40ml 18.05mL Volume of HCl used in Titration, ml1265my 14.65ml Moles HCI .00233 .00 293 00293 Moles Borate ? Molarity of Borate Molarity of Nat К Average K of Borax: of 2mHel! 12.65 ml 1000 how many mores in (0.2 mm) (61265) - 6.00 253 n= CU

Answer 1

Q1 to 4 are regarding calculations from Part B) data. Similar calculations can be done for Part A) as well.

Note that milli moles = 10 -3 x moles

And volume for Sodium ions is also 10.00 mL.

The Molarity for both the trials for Na ions and Borate has been calculated:

cept Borax, Nag By 05 (01)4.8H₂O is spasingly soluble salt, Borax Ź 2Na+ (aq) + BAOG (04) (aq) + 8 4₂0 Borate and Borate Ksp = [Na+] ² [Borate i d tituated using na as: Bosate (aq) + 2 HU + 3H2 w 4 H₂ BO3 + 20 tag) . (aq) for a .. 2 moles of nice n mole of are nee used used a mole Borate 2 mole Borate And or Calculations For Part B. I mole in Borate - » 2 mole of Nat n mole of Nat - & 1. for for moles of He Trial 1 (TI) = Trial 2 (T2) = used = volume used Molasity (5.780.2)x10 s = 10.00114) (709x 0.2) x103 = 10.00158 =10.00158 0.00079 Q2. from © abone, for Ti, Bosate moles = 60.00114) =10.00057 For T2, - = (0.0058) = {0-00079 Q3. From © abone, moles of Nat = moles of Hel. [M) molarity chesday = moles = milli moles 1V=10.00mL) Volume of solution (inL) milli litse for Bosate : (TI); MF, = 0157 - 60-0529] To): MEE 0:29 - 10:079M] For Sodium : (To): M4, = ou 1.4 = 5-714-7)(52): mea auto 1158 - 10.158 M) given 10 10 i

4. So, using equation A,

For T1: Ksp = (0.114)2 x (0.057) = 7.41 x 10-4

For T2: Ksp = (0.158)2 x (0.079) = 19.72 x 10-4

Average Ksp = {(7.41 + 19.72) /2 }x 10-4    = 1 .35 x 10-5