Question

1.Determine the pH during the titration of 36.6 mL of 0.304 M ethylamine (C₂H₅NH₂ , K_b = 4.3×10^-4) by 0.304 M HI at the following points.

(a) Before the addition of any HI

(b) After the addition of 16.1 mL of HI

(c) At the titration midpoint

(d) At the equivalence point

(e) After adding 51.2 mL of HI

b.Determine the pH during the titration of 61.4 mL of 0.450 M nitrous acid (K_a = 4.5×10^-4) by 0.450 M NaOH at the following points.

(a) Before the addition of any NaOH

(b) After the addition of 15.0 mL of NaOH

(c) At the half-equivalence point (the titration midpoint)

(d) At the equivalence point

(e) After the addition of 92.1 mL of NaOH

Answer 1

ats NH + Ho - {H_NH + + oH Il 0.304 -1 0 0 X I tx | tX 0.304-X 1 X [ats NH₃ + ] LOH] [&HENI, k = 4.34104 = 2 0-304-X For weak base 0.304-X E 0.04 4.34104 - 0-304 x= [OH-] = 4.34109 x 0.304 = 0.011433M pOH = -log [OH-] = -log Carol1433) = 1-942 pH = 14-pOH = 14- 1.942 = 12.058

< HsNt molarity = 0.304 M HI: Molanity = 0.304M volume = 36-6m 120.0366L volume = 16.1m/= 0.01612 no ot moles=0.304xcoolel 0.0048944 - CH264 & H NH + HI - SH, NH3 I pH = 14-polt pH = 14-3-265 -10.735 no of meles a 0.304 X0-0366 I .- 264 1 o. co 48944 -0.0048944 -0.0048944 to 00 48944 Elo. 606232 la-00489 44 Total volume =0.0366 to-elblo 0527L 0.006232 -0.11825M ISWWW] 20-05527 0.0048944 0.09 2 87m it mid point pott a pho so pott = ptb = 3.37 pot = 3-37 pH = 14-pelt = 14-3-37=10.63 @ at equivalence Point moles of moles of 0.0ll1 264 & HaN HI ES HSN Hz I] = 0.0527 molarity of H. I - moles of HI volume of HI 0.0111264 0-304 = volume a ti kg = 4-3 4104 pkb -log kb = -10g (4 3 1104) = 3-37 pOH = pkb log [&ltsWHz II. Įsts NHI 0.09887 0.0366 L volume of HI = 3.265 polt= 3-37 +10g 0.11825 -

CH₃NH₂ + weak base ItI - strong acid {H NHI salt o Jo.0111264 1-o-oll1 204 0.011264 -0.011264 to-0111284 0.0 111264 Total volume = 0.0366 + 0.0366 = 0.0732 L [9] Az-N H, I] = = 0.0111264 = 0-152 M _[0] 0.0732 For this salt pH formula lokw=14 pH = 2 [ pk wapkb _logc] = √ [14-3-31-log (0.152)] = 5-724 @ HI Holarity = 0.304M volume = 51-2 ml = 0.0512 L no of moles 0-304 X0-0512 -0.0155648 ASNH + HI - SHINHAI Io.o 111264 16 0.0155648 T-0.0111264 -ooolll 264 0.00 44 384 Toral volumes 0-0366 + 0.0512=0.0878 L 0.0044 384 [HI] > [17] = 7 - 0.050s5m 0 UM ? V -OP7R pH = -log[H+] = -10g (0.05055) = 1.296

6 HNO: Molarity = 0.45m I tx & HND Ho - Het + nor I boles | 0 To 6.45* - x x [Hot] [wog] tvo] NaOH Molanity = 0.45M volumes 61.4 m1 = 0.06142 volume = 15ml = 0.0152 Man not moles = 0.45X0-015 no of mules = 0.458o.obly - 0.00675 0-02763 ANG + Nach - NaNO + H₂o ka 4.5*104 10.02763 10.00675 a plan-199K9= -log (4.5 x164) -0.00675 -0.00 625 +0.00 675 1 -1 pka_3-347 clo.02088 T o 10.oot 75 l Total volume=0.0014 + 0-015 = 0.0164 L 0.02088 CHWqla -0.273314 17.0764 [Nangla 0.00675 20.08835M 4.5x104 = x 0-45-8 0.45 0.45 Forweak aard 4.5810-4- x² . 0.45 1 [Nano] 0.0764 [Nano, ka+109 109 - P"p I IANO] x= [H ct] = / 4.54104 x8-45 [Hot] = 0.01423M PH2 -09 [Hot] = -log (0.01423) pH = 1.847 0.08835 pH = 3-347 tlog 0.2733 pH = 2.8566

☺ At mid pornt pH apka oka = 3-347 so pH = 3-347 @ at equivalence point moles of ANO = moles of 0.0 Nach { Nang} = = 0.02763 = 0.225M = {c} 1 pkw=14 0.1228 For this salt pH formula pH = £ {pk wapka toge] = ± [14+3-347 +10g (0-225)] = 8.35 molarity of waoH moles of Nash volume of Nash - 6 Nach molarity = 0-45M volume = 92.1 ml=0.09212 no of moles = 0.45X0.0921=0.041445 ANO + NagH - NaNg + Ho I 0.02763 10.041445 -0.02763 -0.02703 EL0 -013815 Total volume = 0.0614+ 0.09272 0.1535L 0-0 2763 0.45- - volume of Nash 0-013 F15 20.09M In ac4} = 04] = 7 * 3.1535 polt = -log [OH-] = -log(oroq) = 1.046 pH = 14-pOH = 14-1-046212-954 volume of Naot = 0.0614 L HNO Nash - Nanotto weak acid strongbase salt I 0.02763 10-02763 0 Choo2763 -0.02767] +0.02763 0 0 0.02763 Total volume = 0.0614+ 0.0614 = 0.1228 L

pH formula's for salt's weak acid+ strong base -salft ito pH = ₂ [pkw tpka + logc] strong acid tweakbase salf tho pH = ₃ [ pk w plebe_loge]