Question

HCO3 + HCO3 = Acid Base 3. Complete the table of values below for the aqueous solutions at 25°C using your calculator. (No need to show your calculation here) [H] [OH-1 PH рон 2.7 x 10-8 M M 8.60

Answer 1

Answer At 25°c, het sowbility product of tho is kw Ionic equilibrium reaction of pure water- 2hole thot (aq) + OH (aq). For pure water at 250e concentration of thot and of are same, so [thot ] - [on ]. kwa[hot] [ok] = [hot]2= [OH-]2, » [th of] = Tkw , [ok] = Nkm. At 25°C kr. Kw=10-4 1014 So, so I thot] [OH-] = kw=10-14 I thot 317) [H] = 2.7 X10-8 M. in Kw = [ht] [or] 10-19 = [ht] [or ] [OH-] = 10-19 10-19 [ht] 2.7810-8 COH-] = 3.7x1007 M. = 3.7 x 10-7 M. bHE-log [ht] - Yog (2.7 X10-8) = 7.57. pOH = -log conn] 10g (3.7 X16-7) = 6.43 = 19–pH. 1 por = 8.60 = -log [OH-] >> [OH-] = 10-POH = 10-860 - 2.51X10-9M. kwa[ht] [OH-] = 10-14 [ht] = 10019 - 10-19 [0H 2.51x10-9 . [ht] = 3.98 X10-6 - 3. 98x10-6 M Ha-log [H] = -log (3.98 X166) = 5.40=14-pot-

after Henie, calculation final solution is - Lон] pң рон 2.x [о- 8м .1 х 10- + м 1.5+ 6: 43 3. 92 хос “м 2:51 х 10-9 м Б. 40 8. Бо