Question

A student titrates a 22.0mL solution of 0.200M HCN with 0.200M NaOH The acid disse HCN is 6.2x10-10 Of The acid dissociation constant of 1. Determine the pH of the acetic acid solution before any NaOH is added 2. Determine the pH of the solution when 6.00mL of 0.100M NaOH has been added. 3. Determine the pH of the solution when17 OmL of 0.100M NaOH has been added. 4. Determine the pH of the solution at the equivalence point.

Answer 1

Naon T 1. (0.2M) / The Burette VNoolu - Ome HON = HT + CW, Ke = 612X155/0. 020 Mis to a Titration flask 0.2 -u I t-teq ~0.2 HeN , 22ml (0.2M) [HT] [EN] Chand 6.2X15 C Chopa unity as activities of pure liquid is ccnsidered to nou o.2 s. 6. 2*/olo 42= be unity 1.248 10-10 x2= 12 4 x 10-12 THAIR narizy X 150M = 11. 13x10-6 m) pH = -log Chty = -log, (4-13 X10%) = 4,953 2. Nach - Gune 5 Coup - 6X02 , CHCI, 2 0.242 final concentrations of on a HCN just 29 before the reaction

a humiting reactant HCN Tour CN theo, keq = Ka = 61281570 К. Ab 0.2 X22 0.2 x 6- trt 29 cure In Memoralischen lication 6,2x, je Lisboa 0,2X14 0,206 22 28 0.2816 a 0.226 22 x - n } - Hyelolysis by Conjugate 28 0.2x 20. 2X6 28 28 Base of weak acid re- cã Keq = 6.2X10" = Oxxy3 20xnxo2X 48 2 Cožda to ist eine thens Con}a no 0.06048 8154 pon = -log, (0.0 6048 X 15") = 5.22 i H 214 - pok = 14-5.22=8.78] 3) UNOH 317 me l imiting reagent - MCN Pohl = 20, kes Abbat 22 x 0.2 6.2 X10" 17802 39 39 5X09 178 0.2 3۶ Ab sx ora 14*0,2 ek 19 a 5X 0,2 17X0.2 25

1- 0 39 xnx Restor а 4 : 2x 9 Х4.2 Сол? • 1 x1.-4. oste хот Сов, рок - -•2.19ң)- - Сезо»гче хочу = а. th = I{-poh • (7 - . . . 4, Ab equivalence point, mollier of Hen = milline of Noon (22 хо•2)x 1 т і сок о. 21 [ Nasr = 22 me? истон = см *н, ә , keg 24, 2x71 р, 2x2 p. Ex21 чч 44 О. 2х22 чу * * 0.2x1- - 0-2xt- - ~0.2хгі ५५ ЧЧ Оххх и х9хх, 99/09 Соңлэ 1.2-хоз 2) poh = -2 (1» 27х + 2) 2 ° 2, 244 рі г 14 - рон //p1 ° 11.1он о» | хочах. 1.2. 1, 413xt-6