Question

Calculate the pH of 1.00 L of a 0.244 M hydrocyanic acid solution before and after the addition of 0.455 mol of sodiumcyanide.

pH before addition =
pH after addition =

Answer 1

HEN: Ka= 6.2x10 10 (From google). pka = -log ka = log (62x1000) = 9.21 before addition HENT HO 7 Hz ott en Il 0.244 - olol 1 -X 1- xx xx E10.244-X -1 k [Hot] [cN EHEN 6-2x10 10 = x² -10 0.244-x (-2x1010 10.244x) = x² 1:5128 x 20lo _ 6.2x100 (x)=x² x²76-2x10 10 (X) - 1.5128 x1070 zo ax + bx +c=0 - biſ 6²-490 +(62x1000) 2 - 46116-1.5128 81010) 261) – 6.2x10 10 7 0.0000 244949 - 6-2x10107 / (8.2010 1072 2 - 6.2x10100-0000244949 x= - x=0.0000122471 M: Chot] pH-1092Hot]: -109 (0.0000122471) = 4.912

after addition [F] - - 244 M {Naciv]= = 0.455 20-4554 pka_9.21 - [Nacin] olt - pka tlog EHCN] pH = 9.21+ logo 9 0.455 = 9.48 .244