Question

I LICN is asalt solution. The ka of HCN is 4,2 x 10 - a should the salt solution be auldic, basu, or neutral. o LiGN 7 Li + + CW - LIOH + HEN Basic SB WA b. Write the eauillibrium reaction for theion of the salt that contributes to the pH of the solution LiOH + HON O LICN + H2O Base Acid conjugate conjugate base C. Calcunate the pr of a 1,25m solution of LiCN LiCN t H2o I L1,25 LH с — X E 1.25-Y LiOH + HON 0 0 tx tx

Please check my answers to part a and b, an help me answer part c. I am not sure if I set it up correctly. Show work please, thank you.

Answer 1

a) Li CN is composed of weak acid(HCN) and strong base(LiOH) pH of such salts is more than 7 as they are basic in anture. LiCN (aq)H2O() Li* (ag) CN (aq) +H20() CN (ag) +H20() > HCN (ag)LiOH (a) >HCN(ag) + Li' (ag)+он" (ад) HCN (aq) + OH (aq) Thus, the hydrolysis of LiCN ends up in forming the basic soluiton b) Equation representing the ionizati on of CN is as follows: CN HO HCN OH Acidionization constant K of HCN = 6.2x10-1 CN is the conjugate aci d of base of HCN 1.0x10-14 6.2x10-10 Baseionization constant K, of CN K 1.6x10- -= c) Concentrati on of LiCN 1.25M Equation for the dissociati on of KCN: Li (aq) Since 1 mol of LiCN forms 1mol of CN, CN[LiCN LiCN (aq) CN (aq) 1.25 M CN H2O ICE table: Initi al concentrati on (M) HCN OH 1.25 0 0 Change (M) x x Equilibrium concentration (M) 1.25-x HCNOH CN Baseionizati on constant, K, = 1.6x10- x)(x) 1.25-x 4.47x103 M According to the equilibrium table, OH(x) M 4.47x103 M 1.00x10-14 4.47x 103 According to the ionic product of water, H0* Он" = 2.24x10-M pH-log H0]log (2.24x10-12)= 11.65