Question

Consider the titration of a 25.0mL sample of 0.125M HCN (Ka = 6.17x10^-10) with 0.160 M KOH.

1. What is the initial pH?

2. What is the pH at 6.00 mL base added?

3. What is the pH when 19.53 mL of KOH is added?

4. What is the PH if 21.5 mL of KOH is titrated with the solution?

Answer 1

25ML ef o. 125 M HCN 25m x o125 mel 3125 mmd HCN Ka = 6.17 xl0to Pka-log ka9 2 HCH Ht +CN t-o eqm -Ca Co Col Co2 Ka =(Callca) C:uoncenbialion= D.125 M : dlegre of dissoiaten TRS 1 Ka= Col od 6.17x10 125 Ka CH J=Col=O125)M+. 026 Mo) =8.782 Xlo6M pH-19TH PH: 5.650 5.056 2) 6ml e o.l6oM KoHaddlod nt 6 my xo16 mel 0.96mmol Kok KOH + HCN KCNH,O 3 Agaction Jakig flae 1:1 mole sraho , 0.96 mmol el koH will heaut with 0.96 mmol ef HCN to fordue KCN 0.96mmol nkeA 0.96 mma CKCNJ= KeN 0.96 mol Vto tal -0l =0.0304 M

= 3.125mmol 0.96 m mol = 2.165mmd instrally reactcd CHCNT=2.165mpnc O.0698 M PH=pka+ log [salt]KCN TAidJ HCN + lag 9.21 O.0698 PH = 8.857 19.53mL KoH 0160M 14:53 mL X 016 mol- 3.1248mmo HEA KOH 3 1248 mol 25+ 14.53)AL 0.07017M Tkeu -6 4.491X10M 3.125-3.1248)mo (25t 19.53) mL 1 CHCNJ =

Snce nauid and Cid Vy less hich can gnored js be 3 1248 /PH= 9.7057 Thus,pH at 19.53 mL KoH addite is Frerd found to ka 9.7os

21.5 mレ 0.16 M KoH 2l.Sm x 0.16 mol 3.44 mmol KoH Kot 3.125 mmol will heait with 3.125 mmol HuN to o 3125 Mmol KCN Thus (nKoeea 344-3-125 eecers O.315mmol Total Voluno 25 mL+ 21 Sm 46.5mし COH- 0.315rm 46 SAL 6.7742x10 M -3

POkglo] = -lg 742 53 2117 ph:14-poH 83 ThpH-1.83 KoH, pH Aptai encen is fond to be 83 addilten