Determine the pH at the point in the titration of 40.0 mL of 0.200 M HC H,O, with 0.100 M Sr(OH), after 10.0 mL of the strong base has been added. The value of Ka for НС Н,О, is 1.5х 10%. ( PREV NEXT> 1 2 3 4 Based on the result of the acid-base reaction, set up the lCE table in order to determine the unknown. HC,H,O,(aq)+ H,O(1) = H,O(aq) + C,H,O,(aq) Initial 0.120 0.0400 + x Change +x 0.0400 (М) Equilibri um (M). 0.120 - x +x 0.0400 + x CRESET 0.100 0.120 0.140 0.0200 0.0400 +x | 0.100 + x 0.100 - x 0.120 + x 0.120 - x 0.140 + x 0.140 - x 0.0200 + x -X 0.0200 - x 0.0400 + x 0.0400 - x

Chemistry: Principles and Reactions
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Author:William L. Masterton, Cecile N. Hurley
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Chapter14: Equilibria In Acid-base Solutions
Section: Chapter Questions
Problem 33QAP: A buffer is prepared in which the ratio [ H2PO4 ]/[ HPO42 ]is 3.0. (a) What is the pH of this...
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Question 11 of 14
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Determine the pH at the point in the titration of 40.0 mL of
0.200 M HC,H,o, with 0.100 M Sr(OH), after 10.0 mL of
the strong base has been added. The value of Ka for
НС НО, is 1.5 х 10%.
( PREV
1
3
4
NEXT>
Based on the result of the acid-base reaction, set up the ICE table in
order to determine the unknown.
HC,H,O,(aq)+ H̟O(1) = H,0ʻ(aq) + C̟H,O¿(aq)
Initial
0.120
0.0400 + x
(M)
Change
+x
0.0400
(М)
Equilibri
0.120 - x
+x
0.0400 + x
um (M)
CRESET
0.100
0.120
0.140
0.0200
0.0400
+x
0.100 + x
0.100 - x
0.120 + x
0.120 - x
0.140 + x
0.140 - x
0.0200 + x
0.0200 - x
0.0400 + x
0.0400 - x
+
Transcribed Image Text:Question 11 of 14 Submit Determine the pH at the point in the titration of 40.0 mL of 0.200 M HC,H,o, with 0.100 M Sr(OH), after 10.0 mL of the strong base has been added. The value of Ka for НС НО, is 1.5 х 10%. ( PREV 1 3 4 NEXT> Based on the result of the acid-base reaction, set up the ICE table in order to determine the unknown. HC,H,O,(aq)+ H̟O(1) = H,0ʻ(aq) + C̟H,O¿(aq) Initial 0.120 0.0400 + x (M) Change +x 0.0400 (М) Equilibri 0.120 - x +x 0.0400 + x um (M) CRESET 0.100 0.120 0.140 0.0200 0.0400 +x 0.100 + x 0.100 - x 0.120 + x 0.120 - x 0.140 + x 0.140 - x 0.0200 + x 0.0200 - x 0.0400 + x 0.0400 - x +
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