For the reaction below, Kc = 1.10 × 10-8. What is the equilibrium concentration of OH- if the reaction begins with 0.610 M HONH,? HONH2 (aq) + H2O (I) = HONH3* (aq) + OH¯ (aq)

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For the reaction below, Kc = 1.10 ×
10-8. What is the equilibrium
concentration of OH¯ if the reaction
begins with 0.610 M HONH,?
HONH2 (aq) + H2O (I) = HONH3*
(aq) + OH- (aq)
M
1
2
4
6.
C
7
8
+/-
x 10 0
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LO
Transcribed Image Text:Question 2 of 6 Submit For the reaction below, Kc = 1.10 × 10-8. What is the equilibrium concentration of OH¯ if the reaction begins with 0.610 M HONH,? HONH2 (aq) + H2O (I) = HONH3* (aq) + OH- (aq) M 1 2 4 6. C 7 8 +/- x 10 0 Tap here or pull up for additional resources 3. LO
Expert Solution
Introduction

In equilibrium concepts, it is quite common to come across various equilibrium constants. The equilibrium constant (K) becomes quite helpful while expressing the relationship between products and reactants for a particular reaction (under equilibrium conditions). 

Given data: 

The value of Kc is 1.10 x 10-8 

The initial concentration of HONH2 is 0.610 M. 

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