Question

(a) What is the basis of the approximation that avoids using the quadratic formula to find an equilibrium concentration?

Answer 1

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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