Question

(1) Boric acid sample (H3BO3) contains boron (Na2B407.10H2O) as impurity. The quantity of this sample was dissolved in distilled water to give 1 liter of A solution. (a) 25.00 cm3 of solution A required 8.80 cm3 of 0.01 M HCl solution for neutralizing, using methyl red as indicator. (b) In another 25.00 cm3 of A was added 8.80 cm3 of the 0.01 M HCl solution, and some cm3 of glycerol was added. The result is a uniform colorless solution. The final solution was found to be equivalent to 27.30 cm3 of 0.1M NaOH solution, using phenolphthalein indicator. Find the purity of boric acid.

Answer 1

Answer Given, Boric acid (H₂B03) contains boron (Naz Ba 07 - 10 H₂O) as Solution A is prepared by dissolving the sample into 1 lit. of water. - In water, the Na, B4O; . 10 H₂O forms, basic solution. . NazB407 + nHEO - 2 Nat caq.) + 4 HyBOg + 204 laged For, 250 mL of solution A, 8.80 ml of 0.01(M) HA is required for neutralizing, using methyl red indicator In 8.80 mL of 0.01(M) Hu, moles of Hcl = 8.8x105 moles For, I mole of Nag B407.10H20, 2 moles of NaOH is produced. i moles of Na, B₂O₃ - 10H₂O present in 25 mL of solution A is = 88x108 = 4.4x1ős moles . In another 25 mL of solution A , 8:80 mL of 0.01(M) Hel an some c.c of glycerol was added.. Hence, the reaction takes place, 2 CH 2 OH H₃ BO ₃ + 2 HO + 3H20 HO + 3H ск-он CH₂-OH lo HOJ ΝΑΦΗ 27.3 mL of 0.1(M) untren is required to neutralize the solution. & Moles of NaOH in 27.3 mL of 0.1(M) Naoh solution is = = 0.00273 moles of NAOH 27.3 xo'l 1000 ii In 25 mL of solution A, moles of HyBOg present = 0.00273 Moles of H₂ BO₂ , from Naz BG 07:10 H₂O = 4*4-4x105 = 0.000176 moles i moles of pure boric acid present in 25 mL of solution A = 0.00273 - 0.000 176 = 0.002554 moles . purity of boric acid = 0.002554 = 93.55.1. 100 / 0.00273