Question

10. Determine the hydroxide ion concentration and pH of 0.0942 M hypochlorous acid solu- tion. K = 2.85 x 10 Answer 11. What is the pH of 20.00 mL 0.100 M acetic acid solution that is mixed with 30.00 mL 0.200 M sodium acetate solution? K = 1.8 x 10% Answer 12. Determine the fluoride ion concentration for a saturated solution of magnesium fluoride. K = 7.4 x 10 Answer

Answer 1

4 we know moles of mola bity means solute a dissolved number of per litre of solution.. moles Volume (1) arity = M - : Þau unit of molarity = moles) NOW number of moles of acetic acid in 20 mit 0.100 in acetic and = 0.100 reole 20 = 2x10 3 more and number of moles of sodium acetate in 30mL of 0.200 n CH3 Coona = 0. 200 mole z - 1L x 30 L Toro = 6x10s hole Total volume = (20 mL +30m2) = 50m2 0-052 a. [CH3COOH] in mixture 2 103 and [CH3roo Ha] in min ture = 6x 103 male ka for aceticacid = 9.8x105 tipka s-logka = -log('& *10-5) mole 0.05 L 0.05L According to 4.2 . - equation, According to Henderson- Hasselbalch pH = pka + log [Base => pH = -log ka + log (tcha colla 3) pH = 4.744 +100 P = 5.221 loak ILCH3 Coonal + (o TCH 3 COOH) 6x103 0.05 2x10-3 0.05 The measured jutralue = [5.221

Hypochlorous and is a very wear acid, It dissociates in water very my partially partially, The dissociation equilibroium can be Written as. ICE table Holl (as) + H2O(l) + H2o+ (aq) + ou can Initial.) 0.0942 change - - Equilibrium (0.0942-) - where, n' be the amount of Hoce dissociates. Now, equilibrium constant, ka= [H3ot) [ou] E Hous 7 2.85 x10 8 = 0.094202) ~ + 2.85x108 - 2.68478109 20 compare with this equation, antbr te zo ,na bt 16-yac - 285x108 + J (2:8 17X1078 gra 4x1x122.684X109) 20 2*1 - 28 5X108 + 10362804 & 25.1799x105 [ 43 ot] = [ou ] = nom = 5:18 x 205 m. NOW [ht] [w] = 10-14 [OH- ] = 1014 TH+) 1014 5.18x16 5 [Mt] [OH-] =kw (4+)[#1214 [on) = (1:9 305x10-19 m And pH = -log [H] = -log[5:18 x 105] = 4:2857

o magnesium fluoride (mg F) is a Sparingly soluble sat. The equation for dissolving of magnesium fluoride: Ice table mafa (s) r M g 2 + (as) + 2F (ag Initial: - Change: Equilibrium: - where, wo be the amount of mgf that dissolves or molar solubility. + tan 2n Now, NOW solubility product ksp= [ngar] [F] * 7.4 X10" = xx (2) - * 7.4x10!! = 413 > n = 2.644X1004 ; molar solubility of uy 2.644810-4 Again ngta dissolues to produce one Mg 2+ ion and two fluoride ions. fluoride ion concentration, [F- = (2 x 2 644 810 4 ) & 5.28 x 104 m