Question

Consider a solution formed by mixing 41.0 mL of 0.100 M H₂SO₄, 76.0 mL of 0.100 M HOCl, 25.0 mL of 0.200 M NaOH, 26.0 mL of 0.100 M Ba(OH)₂, and 12.0 mL of 0.150 M KOH. Calculate the pH of this solution. (Refer to this table of K_a values of common monoprotic acids.) (with the correct sig figs)

Answer 1

weak acid Strong acids ta on it so Given solutions are include strong acids, Strong bases, and w a) 41.0 ml o. 100M HSO 4 so → 2H +so [ht) = 2x 0.100M = 0·200M. b) 76.0ml ,0.100M HOCI : Itoch is a weak acid pka = -log ka = -log (3.5x10%) = 7.456 Ka = 3.5 x108 C) 25.0 ml, 0. 200 M Naoit; [OH) = 0.2004. d) 26.0 ml, 0.100M BaOH); Ba(OH)₂ → Bart (OH) = 2x0.100 = 0.200M. © 12.0ml, 0.150 M KOH; OĦ) = 0.150 M. Molarity of a mixture atter neutralisation can be obtained by votre notation findi as follows At milli moles of acid = molarity x Volume (mi) (From strong, Guide = (410ml x 0-200M) * = 8.2 mmol. mmol of ha oH = (25.0ml x 0.200M) +(26.0ml x 0.200M) + (12.0x0.50 =(5 +5.2 +1.8 mmol = 12.0 mmol. Atter neutralisation, solution will be basic as [CH] > [1+") mnmole of oth left = 12-0–8.2) mmol = 3.8 mmol. Total volume mmoles of itt, from Hocl = 750ml x 0.100M (weak acid) = 7.6 mmol. Total volume of the solution = 41+ 76 + 25 + 26 + 12 F 180 me. Hocl toť totoci Hoc) = mmol = 3.5mmad mmol I 7.6 3.8 -3.8 -3.8 I 3.8 [ocil = 3.8 = 0.0211M - E 3.8 o Now, the given solution (atter mixing) is an andere batter which contains weak acid Hoc and its conjugate base OCT. pkat 1=7.456 7TH OCI PH = 7.46] logl=0 BAGH (KOH) (NaOH) - ml 0.0211 M- 3.8 3 180 Coci. och = 7 56 2.) = 7.4 21)