황산 용액의 pH. What is the pH of 0.1 M H2SO4 (sulfuric acid)?

황산 용액의 pH. What is the pH of 0.1 M H2SO4 (sulfuric acid)?

 

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H2SO4 is a diprotic acid. This acid dissociates in two stops:

1st: H2SO4(aq) → H^+(aq) + HSO4^-(aq)

2nd: HSO4^-(aq) ⇌ H^+(aq) + SO4^2-(aq)

 

Because H2SO4 is a strong acid,

this first dissociation is total and the [H+] = [H2SO4] = 0.1 M

 

The bisulphate ion HSO4^- which was formed, is a weak acid

and it only partially dissociates to contribute H^+ ions to the solution:

You have two possible scenarios:

 

1) If the HSO4^- ion did not dissociate at all,

then as above, [H+] = [H2SO4] = 0.1 M and the pH would be:

pH = -log[H+] = -log(0.1) = 1.0

 

2) If the HSO4^- ion could dissociate completely,

then the [H+] contributed from this dissociation would also be:

[H+] = [HSO4-] = 0.1 M,

making a total [H+] = 0.1 M + 0.1 M = 0.2 M

Under these conditions, pH would be:

pH = -log(0.2) = 0.7

 

But the dissociation is some proportion between these two extremes,

and so the pH must lie somewhere between 1.0 and 0.7

 

In order to do this

you require to know about the dissociation constant of an acid, Ka.

The Ka for the HSO4^- ion is 0.012

 

The HSO4^- ion dissociates:

HSO4^-(aq) ⇌ H^+(aq) + SO4^2-(aq)

 

Ka is defined as:

Ka = [H+][SO4^2-] / [HSO4-]

 

Let us call [H+] = X

What do we know:

1) X = [H+] = [SO4^2-], so we can write [H+][SO4^2-] as X^2

2) the concentration of HSO4^- remaining in solution

= the original concentration of [HSO4-] less the amount that dissociated.

That is (0.1 – X)

3) Knowing that Ka = 0.012, we can rewrite the Ka equation as:

0.012 = X^2 / (0.1 - X)

X^2 = 0.012 * (0.1 - X)

X^2 = 0.0012 - 0.012X

Rearrange to form a quadratic:

X^2 + 0.012X - 0.0012 = 0

Solve for X:

X = [H+] = 0.029 M (other root ignore as it is negative)

 

So the total [H+] = 0.1 M (from 1st dissociation)

+ 0.029 M (from 2nd dissociation) = 0.129 M

pH = -log[H+] = -log(0.129) = 0.9

 

 

Answer: pH = 0.9