Question

(7) 15 points An aqueous solution contains 36.00 mmol of sodium cyanide (pK. - 4.800). in 32.0 mL water. Various volumes of strong species are added to the solution. (a) Find the pH when no amount of strong species has been added to the solution. (That is, determine the pH of the original solution). pH (a)=. 3 sig fig (b) Find the pH when 8.00 mL of 4.5 M perchloric acid is added to the solution. pH (b)=. 3 sig fig (c) Find the pH when 10.0 mL of 4.50 M perchloric acid was added to the original solution pH (6) 3 sig fig

Answer 1

Solution @ ph of original solution moles of Nach = 36.0 mmol volume of H2O = 36m2. concentration of Nach = moles of Nach I Volumes c = 36 moned 36 mL C= 1 (mol) (12) can Nach is a salt, following equilibrium will exist in Schition Na CN can) - Natar + Chlag) CN + H20 - HON+ OH Cars pn of this solution is given by pn = 7+ Ž pka + 1 long c where. c= concentration of 1. Nach Scanned with pka of Hen. CamScanner

91 rеи, ри, = Ч: 4 We have pkat pkb = pkw pka + 4.8 = 14 opka = 9.2 Therefore, ph is given by pH = 7+ 1 X9.2 + + log (1) P1 = 7+46 PM = 11.60 pr = 11.6 l three significant figure. 6 - 8 ml of 4.5m Perchloric acid is added. . Huly (perchlogic aud) is a strong and it with react with Nach moles of per chlorie aud = Melexity x volume = 45 mix & ml. = 36 mmol. +36 +36 +36 Nach + Hloy . HON'T Na clou D36 36 -36 . -36 +36 0 0 Since Nachoy is a salt of strong & base and strong acid so it will not hydrolyze. Po of solution will be determined by Hon. Is Scanned with CamScanner

MIN - nt tch concentration of HCN = = moles of NCH Total volume 36 mmal ич L 0.8181 (mod) = HCN 0.8181 4t+ oH 0 0 se z ř 11 ka = = (4+] [ CN] THCNJ X 0.8181-1) ka of HeN = 62016-10 6.2 X1610 = x2 10.818) -x) aisemall, neglecting x; 6.2 X1o lo = materier x = 2.25 x 10's 14+) = x = 2.25 X105 phe -dogla*) = -borg (0.15X009) pH = 4.65 cs Scanned with CamScanner

o lome of 4.5 M Hudy is added. Males of nclou = 10x4.5 m = 45 mmol. - Na Jou t . HON Hooy - No CN 45 36 45-36 36-36 +36 +36 +36 +36 it means there are gomad illoy, +36 mmol Nagoy and 36 mmol HCN. since Naclon is a salt of strong acid and strong buse it will not take part in pH determination HCN is a weak and but in the presence of Strong acid Hyou it will not take part in pH determination, so ph is due to Helly only. Total yolume = 36+10=46mL... moles of Hulu = gmmodo concentration of neoya gnmel 1 46 mL = 0;19565 (More) (n+] = 0. 1955 (milj PA= -log (40) = -167(019565) [ph = 0.708 Answers cs Scanned with CamScanner