Question

A 27.9 mL sample of 0.367 M methylamine, CH3NH2, is titrated with 0.260 M hydrobromic acid. (1) Before the addition of any hydrobromic acid, the pH is (2) After adding 16.8 mL of hydrobromic acid, the pH is (3) At the titration midpoint, the pH is (4) At the equivalence point, the pH is (5) After adding 59.9 mL of hydrobromic acid, the pH is Use the Tables link on the toolbar for any equilibrium constants that are required.

Answer 1

Ch, N1, + H 2 Ch, N4,1 + 35 — 22.9 х 0,372 б. 24он, (o Before addition of any HBr only Base is present so Сси, ми СН3 м » 92, 94 0:3 42 — . 3 3 ч, 2 а, а - - 4 ҳо Сонг) = kx + c (9) = 4xto e340-36- fon] =√1.6ixios (би) = 1, 24 xor- _p?и - - Одон-) роч — — R- 26 х 10- уриа 2 – 11 •24 -ри — род — грн отч — 2 +, 2 ГРИ - 12 + 27, 2 P" - ( 2 + 3+( 0 12:1 от —

Cha Nyt + B6 O cn, Nh2 & His 97.9% 0.367 1618x0.260 10.23 4.36 10:23 – 4,36 0 Basje Buffer 4.36 Ky= 4.4x10-4 plz = -logko pon = pk + log salt - - 3.36 + log 1.2 4.36 = 3.36+ log 4.20 ۔ ۔ ان رہا ہو pkbe 4-log4.4 pkba 4-0.64 pkba 3.36 = 3.36 + logo.74 = 3.36- 0.13 pol = 3.23 ph=14=pon pln =14-3.23 1pn z 10.72 point I . At the midpoint we should equate the equivalent of both compound for finding the volume of HBs. equivalent of CH₂ NHg = equivalent of mise mid -*27.9 X 0.367 E vx 0.260 27.940.367 10.23 - 2x0.260 - 0.52 - 219 67 Ch, Nr₂t the Chynnt & B- 27.9 x0.367 19.17 X0-260 1023 Sill Basic Jo. 23-5:11. 0 5 .11

Г* *** 241 = 30364 long tones - = 3.36 + log fill 3.36+ S2 = 3.36+ logo.gg - 3,36 -0.004 pon = 3.356 ph = 14-pot = 14-3.356 Iph = 10.644 -o at equivalent point both the equivalent of C Compound is same. equivalent of cha Nng = equivalent of U Be. 27.9 x 0.367 Z uxo. 260 - La 27.9 x0.367 10.23 - Ź =39:34 - 0.260 = 0,260 Cha Nhat hBr - Ch Nhat +35 27.9 x0-367 39.34*0.260 10:23 To:23 10:23 10.23. Ech Nht 10.23 - 97.9+ 39.39 - 0.152 67. 24 0 12 97.9 +39.34 - 2 I only Salt is there 4 so hydrolysis

Cha Nn2+ mot eu, nnat tuo 0.152 0,152-u m kue kn = fonses ] a 0,132-* 22 Torty 4.4810-4 0.152 - 4.4 = n² = 0.152 x 10 4 X1on4 -> u , 152 x loto zu Iv0.034 X1070 g a = 0.184X (0-S E [130 + ] phe-log Int] - pn e logo 1184 x 105 pu = 50 log 0.184 = 5+ 0.73 1pn = 5,73 v Cha Nht Be B chann₂ & 4 BS 27.9 x 0.367 59.980.260 10.23 15.57 o 15:57-10.23 10.23 So only strong Aad is present so h) is Come from nort 15.57-10 23 5.39 0.006 = 97.gt sg.g 87.8

z oooo 6 (n+3 = $x10-3 pre - log Cut] pe = -log 6X10 pha 3 O logo 3-0.77 Ipn = 2.22 on = -
all answer is given plz give up thumb for rating.