The approximation is valid because is very small.
Calculation of concentration:
Since
0.85 M 0 0
(0.85-x)M x x
Now the value of x should be
x = 0.0000229
So based on this, the above concentration should be determined.
Now you will solve using the quadratic formula instead of iterations, to show that the same value of x is obtained either way. using the quadratic equation to calculate [h3o+] in 0.00250 m hno2, what are the values of a, b, c and x , where a, b, and c are the coefficients in the quadratic equation ax2+bx+c=0, and x is [h3o+]? recall that ka=4.5×10−4 .
a: 1
b: 4.5x10⁻⁴
c: 1.125x10⁻⁶
[H₃O⁺] = 0.000859M
As HNO₂ is a weak acid, its equilibrium in water is:
HNO₂(aq) + H₂O(l) ⇄ H₃O⁺(aq) + NO₂⁻(aq)
Equilibrium constant, ka, is defined as:
ka = 4.5x10⁻⁴ = [H₃O⁺] [NO₂⁻] / [HNO₂] (1)
Equilibrium concentration of each specie are:
[HNO₂] = 0.00250M - x
[H₃O⁺] = x
[NO₂⁻] = x
Replacing in (1):
4.5x10⁻⁴ = x × x / 0.00250M - x
1.125x10⁻⁶ - 4.5x10⁻⁴x = x²
0 = x² + 4.5x10⁻⁴x - 1.125x10⁻⁶
As the quadratic equation is ax² + bx + c = 0
Coefficients are:
a: 1
b: 4.5x10⁻⁴
c: 1.125x10⁻⁶
Now, solving quadratic equation:
x = -0.0013 → False answer, there is no negative concentrations.
x = 0.000859
As [H₃O⁺] = x; [H₃O⁺] = 0.000859M
To know more about Equilibrium constant
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