(b) Calculate the pH at the equivalence point. (You may continue on the next page if you need more space.)

Chemistry
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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
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ANSWER B= 4.99
Question 9
(a) A 20.00 mL sample of 0.318 M NII (K. = 1.8 x 10) is titrated against 0.475 M HCL. Calculate the pH of
the solution when 10.00 mL of HCI has been added. [3 marks]
0.021 0.3181
= 6.36x10³mcls NH3
0.475 • 0.01
=4.75×20 ³ mels Hcl
4.75X103 HCl
30x103 L
)
= 0.158
H₂O + N +3-7
0.0644
-X
0.0544-X
6-36 X10³-4.75X10³ NH3 = 0.0544M NH3
30×2032
+
•NH₂ TOH
6
0.158
+x
x
0158+X
(b) Calculate the pH at the equivalence point.
(You may continue on the next page if you need more space.)
Kb = [NH4+][OH]
(NH3]
1.8x0² = (0.158+x) X
(0.0544-x)
18 XỐP (0)
0458
(OH-] =
X
pott=-tag 50+]
pott S.20
EX
x = 6.197x10
14-5.20 = 18.79=pH)
Transcribed Image Text:Question 9 (a) A 20.00 mL sample of 0.318 M NII (K. = 1.8 x 10) is titrated against 0.475 M HCL. Calculate the pH of the solution when 10.00 mL of HCI has been added. [3 marks] 0.021 0.3181 = 6.36x10³mcls NH3 0.475 • 0.01 =4.75×20 ³ mels Hcl 4.75X103 HCl 30x103 L ) = 0.158 H₂O + N +3-7 0.0644 -X 0.0544-X 6-36 X10³-4.75X10³ NH3 = 0.0544M NH3 30×2032 + •NH₂ TOH 6 0.158 +x x 0158+X (b) Calculate the pH at the equivalence point. (You may continue on the next page if you need more space.) Kb = [NH4+][OH] (NH3] 1.8x0² = (0.158+x) X (0.0544-x) 18 XỐP (0) 0458 (OH-] = X pott=-tag 50+] pott S.20 EX x = 6.197x10 14-5.20 = 18.79=pH)
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