Calculate the equilibrium concentration of H 3 O H3O in a 0.20 M M solution of oxalic acid. Express your answer to two significant figures and include the appropriate units.

The equilibrium concentration of ion is

Explanation:

We are given:

Molarity of oxalic acid solution = 0.20 M

Oxalic acid is a weak acid and will dissociate 2 hydrogen ions.

Initial:        0.20

At eqllm:    0.20-x                            x                 x

The expression of first equilibrium constant equation follows:

We know that:

Putting values in above equation, we get:

Neglecting the negative value of 'x', because concentration cannot be negative.

So, equilibrium concentration of hydronium ion = x = 0.083 M

Initial:         0.083  

At eqllm:    0.083-y                      0.083+y               y

The expression of second equilibrium constant equation follows:

We know that:

Putting values in above equation, we get:

Neglecting the negative value of 'x', because concentration cannot be negative.

So, equilibrium concentration of hydronium ion = y = 0.0000639 M

Total concentration of hydronium ion = [x + y] = [0.083 + 0.0000639] = 0.0830639 M

Hence, the equilibrium concentration of ion is

in the given question, the partial pressure is calculated by using the ideal gas equation, that is, pv = nrt

here, p is pressure, v is volume = 25 l, n is number of moles = 0.0104 moles, r is gas constant = -0.0821 l.atm / mol.k

t is temperature = 273 k

p = nrt / v

p = (0.0104 mol × 0.0821 l.atm / mol.k × 273 k) / 25 l

= 0.0093 atm

answer by yourhope:

how do models compare to real-world objects?

d. most models represent a real-world object in a simpler way.

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