Question

calculate the pH of 5.0*10^-8 M HCL what fraction of the total H is derived from dissociation of water? use systematic treatment. 1 charge bal

2 mass bal

3chem equilibrium

4 Find pH of solution, step by step please, i am stuck on solving 5*10^-8 + kw/H =H do i use quadratic equation or something?

5 fraction of the total H in the solution derived from water?

Answer 1

1a) Concentration of HCl = 5.0x10M Equation representing the ionization of HC1: HCl(aq) →→ H+ (aq) + C1" (aq) Since 1 mol of HCl forms 1 mol of H* upon dissociation [H*] = [HC1] = 5.0x10M pH = -log[ht] = -log(5.0x10°) = 7.30 The pH of HC1 should be less than 7 since it is a acid. So, we cannot neglect the hydrogen ions furnished by the auto onization of water Equation representing the ionization of water: HO +H* + OH- Ionic product of water, K = [#*H] Ionic product of water,K, = 1.0x10- For a neutral solution, [v+]=[OH] K. = [H](OH] 1.0x10-4 = [H+][v+] 1.0x10-4 = [H7 (H)=1.0x10-'M JHCI VIIVI Concentration of H* ions furnished from HCI, (H* Turn = 5.0x10M Concentration of H+ ions furnished from water, [H* L = 1.0x10-PM Since [H* Tuner > [H*]ck, we cannot neglect the concentration of the hydrogen ions furnished from ther ionization of water. Total concentration of H+, [H*] = [**]sc + [H+ loks = 5.0x10M + 1.0x10-'M = 0.50x10-?M +1.0x10-?M = 10-(0.50 +1.0) = 1.50x10? Now, pH = -log[*] = -log(1.50x10") = 6.82 b) Concentration of H+ ions furnished from Ha, Hl. = 5.0x10M Concentration of H* ions furnished from water, [H 2 = 1.0x10-'M Total concentration of H", [H*)noktin = [H* ]]c +[* Drake = 5.0x10M + 1.0x10-'M = 0.50x10-M + 1.0x10-?M = 10-(0.50 +1.0) = 1.50x10-7 (H* ]uci Fraction of H* ions from HCl in the the solution = TH* horti 5.0x10- 1.50 x10-7 = 0.333