4. The titration of 25.0 ml of 0.100 M HCI is made with 0.100 M of NAOH as the titrant.a) What is the original pH before adding the base?[H +) - 2.5x(0- x 0.1=2.5X10-4 M2.5 x 103pM=-10g 25KCO-3)-PH= 3-10g25PM=2.60b) What is the pH after adding 5.00 ml of the base?2.0- 0. O66 MpH= - log (G066)PH=1.18CH+]- N,V ーNV230Vi+VZ(2.5 x0.1)-(5x Q1)2ら+5c) What is the pH at the equivalence point?Ateguivalen ce point, neutrilization takes place chen PH =73ラX 0.1 -Vx 0.1V 25 mLd) What is the pH at half the equivalence point?At half eguivalence point, only half mules of HCI Pet out in 42.5xi0-3# of moles of acid left:- (uncentra tin of ucl= 1.25o3=1.25 O-3%3DO0333PH=-1og(0.0333)e) What is the pH affer adding 30.00 ml of the base?PH=- log H+](25 +125) x103pu=1.48f) Sketch the titration curve.

4. The titration of 25.0 ml of 0.100 M HCl is made with 0.100 M of NaOH as the titrant. pH=-log CH] ア- PH= 3-l0g5 PM=2.60 a) What is the original pH before adding the base? [H+) - 2.5x10-3 x 0.1 = 2.5X10-4 M 2.5 x 10 3 pM=-log 2.5K(03) b) What is the pH after adding 5.00 ml of the base? 2.0 [4+]= = O.066 M pH= - log (G066) PH= 1.18 CH+]- N,V, ーNV2 Vi+VL (2.5 x0.1) - (5x G1) 2ら+5 c) What is the pH at the equivalence point? At eguivalen ce point, neutrilization tahes place hen PM =7 3ラX 0.1 =Vx 0,1 V 25 mL d) What is the pH at half the equivalence point? At half eguvalence point, only half moles of HCI Pet out in t + of mules of acia lef :- Cuncentra ton of ucl= 1.25 o3 P4 = - log ] 25xi0-3 =(125 N0-3 %3D O0333 (25 +125) x3 Pu=1.48 PH -log (0.0333) e) What is the pH affer adding 30.00 ml of the base? f) Sketch the titration curve.

4. The titration of 25.0 ml of 0.100 M HCl is made with 0.100 M of NaOH as the titrant. pH=-log CH] ア- PH= 3-l0g5 PM=2.60 a) What is the original pH before adding the base? [H+) - 2.5x10-3 x 0.1 = 2.5X10-4 M 2.5 x 10 3 pM=-log 2.5K(03) b) What is the pH after adding 5.00 ml of the base? 2.0 [4+]= = O.066 M pH= - log (G066) PH= 1.18 CH+]- N,V, ーNV2 Vi+VL (2.5 x0.1) - (5x G1) 2ら+5 c) What is the pH at the equivalence point? At eguivalen ce point, neutrilization tahes place hen PM =7 3ラX 0.1 =Vx 0,1 V 25 mL d) What is the pH at half the equivalence point? At half eguvalence point, only half moles of HCI Pet out in t + of mules of acia lef :- Cuncentra ton of ucl= 1.25 o3 P4 = - log ] 25xi0-3 =(125 N0-3 %3D O0333 (25 +125) x3 Pu=1.48 PH -log (0.0333) e) What is the pH affer adding 30.00 ml of the base? f) Sketch the titration curve.

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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