Question

Calculate the pH of a solution that is prepared by dissolving 0.297 mol of pentafluorobenzoic acid (CF.COOH, K, -3.30x10-2) and 0.650 mol of hydrocyanic acid (HCN, K, = 6.17x10-10) in water and diluting to 5.90 L. Also, calculate the equilibrium concentrations of C F COOH, C F C00", HCN, and CN. Do not make an approximation unless the initial acid concentration is greater than 1000 × K. (Hint: The pH will be determined by the stronger acid of this pair.) pH = [C6F3COOH) = M [CF3C00"]= [HCN]= [CN] =

Answer 1

Solution As per given in the question. moles og CFCool = 0,297 kaceescoOH) = 3.3x10-2 moles of It CN dissolved= 0.650 mole Ka CHIN) = 6:17X1070 Yolume of solution = s.gol [ccfs cook] - moleso = 0;297 = 0·05OM moles Volume 0.297 - --0050M sigo (HAN)= 0.660 0.11 M Sigo The kachan) << Kacafroon and concentration of both are in same lange this pH mainly will depends on Cats cool GFS Cool Z ÇEscootha کون ہے ا ater Oros- n an Kalofswh) 10.054) - = 3:3x16-2 ne-b7a -bt T6²-yac ga 2 or n²=1065x103_ 3:3x10² or n2 +3,3x15²2-1:65x163 0 Solving na -33418-2 +/3-34153)+48121.658102 1 2.738162 pha logatta - log 2.73x16² = 1.564 ICFS COORD z Oros010273 = 0.0227M. sfs.com-] 2 U2 0.0273m

HCN is very weak acil it will ionise as - HENZ CN-tut but it = from Gfwon 0.0273M HON Z CN-t H+ 0.1174 0 2737 ate 0:11-1a a 0,0273+. but 0.0273 >>e,': So 010273+4 ~ 0.0273 and onll-ia a orll Kadin) = ax 0.0273 - nx 00273 (oill-Q) 0.11 6117x10-loxoll a nkul 0-0273 • (CN)= 2,488109 m (1 CNY 0111-21488609 0'11M