Question

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3. Consider the weak base-strong acid titration of 25.00 mL of 0.100 M NH, with 0.100 M HCl (weak base, strong acid). The Ks for this weak base is 1.8 x 10 a. Calculate the pH of the solution after the addition of 10.0 mL HCL. b. What is the pH half-way to the equivalence point? c. Calculate the pH at the equivalence point. d. Calculate the pH after the addition of 35.0 mL HCL.

Answer 1

Then pH = 14-pelt pkb -10961-84105) @ Nh molarity = 0 LM Hel: volume = o-lM. 16 at half-way to the equivalence point volume = 25m1=0.0252 volume = loml=0.01L me 2 plaaph no of moleso - 1X0-01 no-ot moles a 0.1X0-025 I = 0.001 so pka = 9.255 į pit = 9.255 2o.0025 con at mid-point from google NH Ho - NH 1 kg =18x105 pott = pkb plaba-log kb I10.0025 0.0010 -o.oo -0.001 to out pkb=4.745 © at equivalence point Eo.0015 00-001 moles of NH = Hel=0.0025 moles of Total volumen 0-025+ 0.01 pa moles of ty : 0.0352 molarity of ka_14-4.745= 9.255 HCl = volume of Hal INH₂ -0.04286M 0.0025 -M INH, ci] = 6. Bol = 0.02857M volume of Hal Is [Nity cl kb [NHA (1) pott apky + log Inte] =p tlog [ NH] volume of Hel=0.025L polt_ 4.7 0.02857 -4.57 Io.04286 pH = 14-pott = 14 + 4.57= 9-43 pka_ 14-pkb 0.035

© Hel: Molarity = o.LM volume= 35m1=0.0352 no it moles= 0.18 0.035=0.0035 - Helt NH3 - Nitq el HC NH3 - NH₂ cl strong acid weakbase salt I 0.0025 0.0025 o T-0.0025 -0.0025 +0.0025 O-0025 Total volume=0.025 0.025 0.05L 0.0025 -0.05 m= {c] 2Nhci- v 0 .05 pkw=14 INICI=1 0.0035 0.0025 lo cl-0.0025 -0.0025 0.000 Total volume = 0-085+ 0-635-0.06L pH formula salt For this 0.001 - 0. OLEETM pH = {p kW _pkb_loge] pH = 7 [14-4-745-10g (6.5)] pH = 4.178 [Hel] = [H +3 = 2 = 0.or pH = -log[4+3 = -log (0.01667) = pita 1-778

pH formula's for salt's weak acid + strong base salft ito pH = ₃ [pfw + pka + loge] strongacid & weakbase salf t pt = ₃ [ pk w _pkebe_loge] o