с. 2.50 х 10 2 МНСIO, OJOH"] = 2.50 x 10-² M, H,O* = 2.50 x 10-2 M OJOH"] = 4.00 x 10-18 M, H,O* = 2.50 x 10-2 M o JOH"] = 2.50 x 10-² M, H3O+ = 4.00 × 10-18 M o JOH") = 4.00 x 10-18 M, H,O* = 4.00 x 10-18 M 4.00 x 10-13 М d. 0.860 M НCI o[OH"] = 0.860 M, H3O+ = 0.860 M O[OH¯] = 1.16 M, H3O* = 0.860 M 0Он- 1.16 х 10 14 М, |н,0*] %-D 0.860 M M, H3O*] оон — 0.86о м, |н,о+ — 1.16 х 1014 М

Chemistry
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Chapter1: Chemical Foundations
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Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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с. 2.50 х 10 2 МНСIO,
OJOH"] = 2.50 x 10-² M, H,O* = 2.50 x 10-2 M
OJOH"] = 4.00 x 10-18 M, H,O* = 2.50 x 10-2 M
o JOH"] = 2.50 x 10-² M, H3O+ = 4.00 × 10-18 M
o JOH") = 4.00 x 10-18 M, H,O* = 4.00 x 10-18 M
4.00 x 10-13 М
d. 0.860 M НCI
o[OH"] = 0.860 M, H3O+ = 0.860 M
O[OH¯] = 1.16 M, H3O* = 0.860 M
0Он- 1.16 х 10 14 М, |н,0*] %-D 0.860 M
M, H3O*]
оон — 0.86о м, |н,о+ — 1.16 х 1014 М
Transcribed Image Text:с. 2.50 х 10 2 МНСIO, OJOH"] = 2.50 x 10-² M, H,O* = 2.50 x 10-2 M OJOH"] = 4.00 x 10-18 M, H,O* = 2.50 x 10-2 M o JOH"] = 2.50 x 10-² M, H3O+ = 4.00 × 10-18 M o JOH") = 4.00 x 10-18 M, H,O* = 4.00 x 10-18 M 4.00 x 10-13 М d. 0.860 M НCI o[OH"] = 0.860 M, H3O+ = 0.860 M O[OH¯] = 1.16 M, H3O* = 0.860 M 0Он- 1.16 х 10 14 М, |н,0*] %-D 0.860 M M, H3O*] оон — 0.86о м, |н,о+ — 1.16 х 1014 М
Expert Solution
Step 1

c) HClO4 is a strong acid which dissociates completely as per the reaction 

HClO4 + H2O --------> H3O+ + ClO4- 

Hence the concentration of H3O+ ions = [H3O+ ] = concentration of HClO4 = 2.50 X 10-2 M

Since pOH = 14 - pH 

=> -log[OH-] = 14 - (-log[H3O+ ] )

Hence substituting the values we get 

=> -log[OH-] = 14 + log(2.50 X 10-2 ) = 12.4

=> [OH-] = concentration of OH- ions = 10-12.4 = 4 X 10-13 M

Hence the answer is option 2.

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