Question

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18. Calculate the pH of a buffer solution that is 0.410 M in HOCI and 0.050M in NaOCI. [Ka(HOCI) 3.2 x 10* a. 0.39 d. 749 b. 3.94 e) 6.58 19. Which one of these equations represents the reaction of a weak acid with a weak base! a. H'(aq) + OH (aq)- H:O(aq) b. H'(aq)+ CHsNH2(aq)- CHSNH3 (aq) c. OH (aq) + HCN(aq)- H:O(aq) + CN (aq) d) HCN(aq) + CH NH: (aq)-CHsNHs"(aq) + CN (aq) 20. Calculate the percent ionization of cyanic acid, Ka 2.0 x 10, in a buffer solution that is 0.50 M HCNO and 0.10 M NaCNO. a0.02 % b. 0.10 % .20 % d. 2.0 % c. Longer Answer (worth 6 points each) l. Calculate the concentration of the oxalate ion (C20in a0.175 M solution of oxalic acid (CH:O4). [For oxalic acid, Kai6.5 x 10, Ka 6.1 x 101 b)6.1 x 105 M a. 0.11 M c. 4.0x 10 M d. 0.0791 M t GHO . רו 6IX105 t (75 G.IAIG 175-r 52 What mass of sodium cyanide must be added to 250. mL of water in order to obtain a solution .07X( +x2 X 17Sx having a pH of 10.50? [Ka(HCN) 4.9 x 10-10 a. 240 g b. 0.032 g c. 0.059 g d. 0.24 g A3.1bx |6" th0 0 4 CIN YqKU10(3.10A1U" (9-516 X10-") -1.SSX10209.94sx1 1.91010 -22 -11 y 3.4 X10

Answer 1

18) The pH is calculated:

pK = - log Ka + log [NaCl] / [HOCl] = - log 3.2x10 ^ -8 + log (0.05 / 0.41) = 6.58

19) The reaction of a weak acid and a weak base is d.

51) The concentration of the second ion in solution is equal to Ka2 = 6.1x10 ^ -5

52) The pOH and [OH-] are calculated:

pOH = 14 - 10.5 = 3.5

[OH-] = 10 ^ -3.5 = 3.2x10 ^ -4

You have Kb:

Kb = Kw / Ka = 10 ^ -14 / 4.9x10 ^ -10 = 2x10-5

The expression of Kb has:

Kb = [OH-] * [HCN] / [CN-]

2x10 ^ -5 = (3.2x10 ^ -4) ^ 2 / [CN-]

It clears [CN-] = 0.005 M

The mass of NaCN is calculated:

m NaCN = M * V * MM = 0.005 M * 0.25 L * 49 g / mol = 0.059 g

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