Question

please the hole steps

Review Problem 16.099 For hydrocyanic acid, HCN, k, = 4.9 10-10 x Incorrect. Did you use the Ky value and not Ky? Did you set up your equilibrium expression in the correct way? Calculate the pH of 0.20 M NaCN. 5.004 pH = the tolerance is +/-2% X Incorrect. The concentration of OH must be equal to the concentration of HCN formed. Look at your work in Part a. What is the concentration of HCN in the solution? 0.00000989 [HCN] = the tolerance is +/-296

Answer 1

Relationship between ka & kb ism Kackg= kw=104 - kb= 10-14 10-14 ka 4.9x10-10 {as ginen. Ka of HCM = 4.8 710-00} 7 kg = 2.0408 xious How, Nacho - Matt Chi - Carth20 kb, HCH + OH- → Kb = [H Che] [OH-] = 2,0408 x10-57 [em] [H20] As, H₂O is the solvent and so considered in excess so. [H20] =1 and [HChi] = [OH-] => [OH-]2 = 2,0408810-5 - [CA ] - → (given [Car] = 0.20 M) 7 [GH] ²= 2:0908x10-5, 0102 = [OH-] = V 410816 xlong * 1[M] = 2.02027 10 * = [hche] 7 poite - larg (22.02 X10-3) = 2.69 458 7 pH = 14- pOH = 14-269458 pH = 11.3057 TpH = 11.311 + [+(n)= 2,01*10*3 2.0, 10