= 0.01 x 0.100 = 0.00100150.0 X0-250 =0.01250.0123025 0.060.300 x0.010.100. X 0.01B. 9.250.066A 10.0 mL solution of 0.300 M NH3 is titrated with 0.100 M HCI. Calculate the pH after the addition of10.0 mL of acid.A. 9.550.001-0.0030.300 x 0.00.001C. 8.950.02PH = PKG +10g [0.01562.67=4.744+ [19165TI-84 PlusTEXAS INSTRUMENTSD. 9.73= Pl= 3.68 - 26728.3.0:30002 0.002 M Plt = -log= .05U.2E. 8.77Pit=Pkg tog20.015= 1.01[A][HA]

Here we are required to find the pH of the solution.

Answered: =1.01 AA10.0 mL solution of 0.300 M NH3…

=1.01 AA10.0 mL solution of 0.300 M NH3 is titrated with 0.100 M HCI. Calculate the pH after the addition of 10.0 mL of acid. A. 9.55 B. 9.25 C. 8.95 D. 9.73 E. 8.77

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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Concept explainers

Acid Base Titration

An acid-base titration is the method of quantitative analysis for determining the concentration of an acid and base by exactly neutralizing it with a standard solution of base or acid having known concentration.

Question

= 0.01 x 0.100 = 0.001
001
50.0 X0-250 =
0.0125
0.0123
025 0.06
0.300 x0.01
0.100. X 0.01
B. 9.25
0.06
6A 10.0 mL solution of 0.300 M NH3 is titrated with 0.100 M HCI. Calculate the pH after the addition of
10.0 mL of acid.
A. 9.55
0.001
-0.003
0.300 x 0.0
0.001
C. 8.95
0.02
PH = PKG +10g [0.0156
2.67=4.744+ [19165
TI-84 Plus
TEXAS INSTRUMENTS
D. 9.73
= Pl= 3.68 - 267
28.3.
0:3
0002 0.002 M Plt = -log
= .05
U.2
E. 8.77
Pit=Pkg tog
20.015
= 1.01
[A]
[HA]

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