We can classify acids by the number of protons per molecule that they can give up in a reaction. Acids such as (ceHCl), (ceHNO3), and (ceHCN) that contain one ionizable hydrogen atom in each molecule are called monoprotic acids. Their reactions with water are:

Even though it contains four hydrogen atoms, acetic acid, (ceCH3CO2H), is also monoprotic because only the hydrogen atom from the carboxyl group ((ce-COOH)) reacts with bases:

*
api/deki/files/126407/imageedit_7_4808614104.png?revision=1" />

1. First Ionization: Determine the concentrations of (ceH3O+) and (ceHCO3-).

You are watching: Write chemical equations for second ionization step of carbonic acid.

Since efstep1 is has a much bigger (K_a1=4.3×10^−7) than (K_a2=4.7×10^−11) for efstep2, we can safely ignore the second ionization step and focus only on the first step (but address it in next part of problem).

As for the ionization of any other weak acid:

*

An abbreviated table of changes and concentrations shows:

ICE Table(ceH2CO3(aq))(ceH2O(l))( ceH3O+(aq) )(ceHCO3-(aq))
Initial (M) (0.033 :M) - (0) (0)
Change (M) (- x) - (+x) (+x)
Equilibrium (M) (0.033 :M - x) - (x) ( x)

Substituting the equilibrium concentrations into the equilibrium constant gives us:

=dfrac(x)(x)0.033−x=4.3×10^−7 onumber>

Solving the preceding equation making our standard assumptions gives:

Thus:

=0.033:M onumber>

=ce=1.2×10^−4:M onumber>

2. Second Ionization: Determine the concentration of (CO_3^2-) in a solution at equilibrium.

Since the efstep1 is has a much bigger (K_a) than efstep2, we can the equilibrium conditions calculated from first part of example as the initial conditions for an ICER Table for the efstep2:

ICE Table(ceHCO3-(aq))(ceH2O(l))( ceH3O+(aq) )(ceCO3^2-(aq))
Initial (M) (1.2×10^−4:M) - (1.2×10^−4:M) (0)
Change (M) (- y) - (+y) (+y)
Equilibrium (M) (1.2×10^−4:M - y) - (1.2×10^−4:M + y) ( y)

< eginalign* K_ceHCO3-&=cedfrac \<4pt> &=dfrac(1.2×10^−4:M + y) (y)(1.2×10^−4:M - y) endalign*>

To avoid solving a quadratic equation, we can assume (y ll 1.2×10^−4:M ) so

Rearranging to solve for (y)

<=y approx 4.7×10^−11 onumber>

To summarize:

In part 1 of this example, we found that the (ceH2CO3) in a 0.033-M solution ionizes slightly and at equilibrium ( = 0.033, M), ( = 1.2 × 10^−4), and (ce=1.2×10^−4:M). In part 2, we determined that (ce=5.6×10^−11:M).


Exercise (PageIndex2): Hydrogen Sulfide

The concentration of (H_2S) in a saturated aqueous solution at room temperature is approximately 0.1 M. Calculate (ce), (ce), and (ce) in the solution:

Answer

( = 0.1 M), (ce = = 0.0001, M), ( = 1 × 10^−19, M)

We note that the concentration of the sulfide ion is the same as Ka2. This is due to the fact that each subsequent dissociation occurs to a lesser degree (as acid gets weaker).


Triprotic Acids

A triprotic acid is an acid that has three dissociable protons that undergo stepwise ionization: Phosphoric acid is a typical example:

The first ionization is

with (K_ce a1=7.5×10^−3 ).

The second ionization is

with ( K_ce a2=6.2×10^−8 ).

The third ionization is

with ( K_ce a3=4.2×10^−13 ).

See more: How Do You Say Years In French Translation Of “Year”, Years In French

As with the diprotic acids, the differences in the ionization constants of these reactions tell us that in each successive step the degree of ionization is significantly weaker. This is a general characteristic of polyprotic acids and successive ionization constants often differ by a factor of about 105 to 106. This set of three dissociation reactions may appear to make calculations of equilibrium concentrations in a solution of H3PO4 complicated. However, because the successive ionization constants differ by a factor of 105 to 106, the calculations can be broken down into a series of parts similar to those for diprotic acids.

Polyprotic bases can accept more than one hydrogen ion in solution. The carbonate ion is an example of a diprotic base, since it can accept up to two protons. Solutions of alkali metal carbonates are quite alkaline, due to the reactions: