Answer: The Concentration of [tex]OH^-[/tex] at equilibrium is 0.000092 M
Explanation:
Equilibrium constant is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric ratios. It is expressed as [tex]K_c[/tex]
[tex]HONH_2(aq)+H_2O(l)\rightleftharpoons HONH_3^+(aq)+OH^_(aq)[/tex]
Initial conc. 0.770 M 0 M 0 M
At eqm. conc. (0.770-x) M (x) M (x) M
The expression for equilibrium constant for this reaction will be,
[tex]K_c=\frac{[HONH_3^+]\times [OH^-]}{[HONH_2]}[/tex]
Now put all the given values in this expression, we get :
[tex]1.10\times 10^{-8}=\frac{(x)\times (x)}{(0.770-x)}[/tex]
By solving the term 'x', we get :
x = 0.000092 M
Concentration of [tex]OH^-[/tex] at equilibrium= (x) M = 0.000092 M
The equilibrium concentration of hydroxide ion is 0.00009 M.
The equation of the reaction is;
HONH₂ (aq) + H₂O (l) ⇌ HONH₃⁺ (aq) + OH⁻ (aq)
I 0.770 0 0
C -x +x +x
E 0.770 - x x x
Note that water is present in large excess
Given that Kc = 1.10 × 10⁻⁸
1.10 × 10⁻⁸ = [HONH₃⁺] [OH⁻]/[ HONH₂ ]
1.10 × 10⁻⁸ = [x] [x]/ 0.770 - x
1.10 × 10⁻⁸(0.770 - x) = x^2
8.27 × 10^-9 - 1.10 × 10^⁻8x = x^2
x^2 + 1.10 × 10^-8x - 8.27 × 10^-9 = 0
x=0.00009
Hence, the equilibrium concentration of hydroxide ion is 0.00009 M.
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