Concentration of H+

Question

HA is a weak, monoprotic acid that dissociated according to the equation:

HA (aq) ßà H⁺ (aq) + A^- (aq)                  K_a = 4.0 x 10^-9

If the initial concentration of HA is 0.10 M, find the concentration of H⁺

Answer 1

Hello Mikaela,

This is an RICE table problem!

First lets start with our Reaction: HA → H⁺ + A^-

Plug in the Initial Concentrations: 0.10M 0M 0M

*If you are not explicitly given an initial concentration you can assume it's just going to be 0.

Now plug in the Change -x +x +x

*Because the leading coefficient of every compound is just one, we can assume we're losing and gaining one x.

Find the Equilibrium concentrations: 0.10-x x x

Now we know that K_a = [products]/[reactants] at equilibrium so we have

K_a = x • x = 4.0 x 10^-9 (given to you)

0.10-x

Now we can solve for x!

x² = 4.0 x 10^-9 => x² = 4.0 x 10^-9 (0.10-x) => x² = 4.0 x 10^-11 - 4.0 x 10^-9 x

0.10-x

This is a quadratic so let's get everything to one side so we can set it equal to zero. I am going to subtract 4.0 x 10^-11 - 4.0 x 10^-9 x from both sides:

x² - 4.0 x 10^-11 + 4.0 x 10^-9 x

and then rewrite this in standard form (ax² + bx +c±):

x² + 4.0 x 10^-9 x - 4.0 x 10^-11

We can now use the quadratic formula: -b ± (√b² - 4ac) ÷ 2a

So we have -4.0 x 10^-9 ± [√(4.0 x 10^-9)² - (4 • 1 • - 4.0 x 10^-11)] /2

You're only looking for the positive value for x since a concentration cannot be negative.

Therefore x, which is the value for H⁺ = 6.32 x 10^-6 M

Answer 2

As above, the ICE table is great:

HA (aq) + H₂O (l) --> H₃O⁺ (aq) + A^- (aq)

Initial 0.1 --- 0 0

Change -x ---- +x +x

Equilibrium 0.1-x ---- x x

Expression for the equilibrium constant:

K = [H₃O⁺][A^-] / [HA]

substitute the values into the expresion:

K = x * x / (0.1-x) = 4.0*10^-9

Now at this stage you COULD go to the trouble of using the quadratic formula.

OR

You could use an approximation and save yourself several minutes of work.

The REASON you can use an approximation is because the initial concentration, 0.1 M, is at least 10,000x larger than the equilibrium constant 0.000000004. When the two values are that far apart, then you can use the approximation. This is because the value of x is SO TINY compared to 0.1 that it won't impact your answer, in terms of signification figures.

I'll prove it:

1) USe the approximation.

x * x / (0.1-x) = 4.0*10^-9

Becomes

x * x / (0.1) = 4.0*10^-9

which is super easy to solve:

x² = 4.0 * 10^-10

x = 2.0 * 10^-5

job done!

2) Use the quadratic equation:

x * x / (0.1-x) = 4.0*10^-9

becomes

x² = 4.0*10^-9 * (0.1-x)

x² = 4.0*10^-10 - 4.0*10^-9 x

x² + 4.0*10^-9 x - 4.0*10^-10 = 0

I plug that into a quadratic equation solver, and the two solutions are:

-0.0000200020001

0.0000199980001

Concentrations are never negative, so just use the positive answer:

0.0000199980001

which, with 2 significant figures is.... .000020 or 2.0*10^-5

Same answer, just a lot more painful!