pH level of 0.5M Nitrous Acid(HNO2,Ka=4.0 * 10^-14)?

Question

i know that Ka= -ve means weak acid so it doesnt dissociate completely. So how do you find pH level? -log()

Answer 1

When you have weak electrolytes, (equillibria in general) you should make the following table.

1. write the reaction

HNO2 <=> H+ + NO2-

2.Put the initial amounts (expressed as concentration) of each molecule. In this case we have only the concentration of HNO2, so

.. .. .. .. .. .. HNO2 <=> H+ + NO2-
Initial .. .. .. .. 0.5

3.Assume that an unknown amount "x" of the reactants react.
Then write also how much of each product you expect to be produced according to the stoichiometry. In this case 1 mole of HNO2 gives one mole of H+ and NO2-, so x mole/L HNO2 will give x mole/L of each and our table becomes

.. .. .. .. .. .. HNO2 <=> H+ + NO2-
Initial .. .. .. .. 0.5
Dissociate .. ..x
Produce .. .. .. .. .. .. .. ..x .. .. .x

4. Ka is an expression for the concentrations at equilibrium. So let's see what are those. For the reactants you will have initial-what reacted. We have only HNO2 with 0.5-x. For the products it will be the initial+what was produced. We didn't have any intial so it is only what was produced (x for both of the products). Thus the table in its final form is

.. .. .. .. .. .. HNO2 <=> H+ + NO2-
Initial .. .. .. .. 0.5
Dissociate .. ..x
Produce .. .. .. .. .. .. .. ..x .. .. .x
At Equil. .. .. 0.5-x .. .. .. x .. .. .x

Ka= [H+][NO2-]/[HNO2] = x^2/(0.5-x)
Now you either solve the quadratic equation and find x, or we assume that 0.5 >>x and practically 0.5-x=0.5.
Then Ka=x^2/0.5=4*10^-14 =>
x=squareroot(2*10^-14)= 1.4*10^-7. It is much smaller than 0.5 so our assumption is valid (otherwise we would have to solve the quadratic and get an exact solution).

For a simple level of chemistry we would now say
pH=-log[H+]= -logx =-log(1.4*10^-7) = 6.85

However for more advanced and accurate expectations we see that the H+ coming from HNO2 is found to be in the same order of magnitude as that coming from the self dissociation of water (10^-7 for distilled water). Therefore in this case we cannot ignore the self dissociation of water (as we usually do) for calculating the pH.
There are 2 approaches for this case, here is one of them.
Remember that we didn't put an initial amount for H+ since in most cases we ignore the self dissociation of water. Now we will consider that we had pure water, so the initial concentration of H+ is 10^-7 and we want to see how much the HNO2 will dissociate under these conditions so our table becomes

.. .. .. .. .. .. HNO2 <=> H+ + NO2-
Initial .. .. .. .. 0.5 .. .. .. 10^-7
Dissociate .. ..x
Produce .. .. .. .. .. .. .. ..x .. .. .x
At Equil. .. .. 0.5-x .. 10^-7+x. .x

and Ka=(10^-7+x)x/(0.5-x)
Now we have to solve the quadratic so we can't do any simplification

0.5Ka-Kax=10^-7x +x^2 =>
x^2+ (10^-7+Ka)x-0.5Ka=0

The only acceptable solution is the positive one
x=0.5(- (10^-7+Ka)+ squareroot (D))
where D=(10^-7+Ka)^2-4*(-0.5Ka)=
=(10^-7+4*10^-14)^2-4*(-0.5*4*10^-14)
=9*10^-14

and x=0.5*(-((10^-7))+4*10^-14) +3*10^-7) =10^-7

so now pH=-log(10^-7+x) =-log(2*10^-7) = 6.70

Answer 2

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RE:
pH level of 0.5M Nitrous Acid(HNO2,Ka=4.0 * 10^-14)?
i know that Ka= -ve means weak acid so it doesnt dissociate completely. So how do you find pH level? -log()

Answer 3

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Answer 4

HNO2 = H+ + NO2-

let x be the moles of HNO2 that reacts in the equation above
we have Ka = x^2 / (0.5-x)
solve the equation we have x = 1.414*10^-7
the pH level = - log ([H+]) = 6.85

Answer 5

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