the H3O+ concentration in an 0.0001
Example: Consider the process by which we would calculate the
Solving this approximate equation gives a value for C
This state originates when the speed of the direct reaction equals the sp… here to check your answer to Practice Problem 4, Click
We then compare the initial reaction quotient (Qa)
Substituting this approximation into the
HOCl, Ka = 2.9 x 10-8, Click
There are two ways out of this difficulty. The concentration of the H3O+
C
OAc-, and HOAc. concentrations agrees with the value of Ka
the H3O+ ion concentration and the
Substituting this information into the acid-dissociation
if Ka is 2.1 x 10-12
these concentrations into the expression for Ka. for this problem. C,
here to check your answer to Practice Problem 2, Click
close enough to 1 to make us suspicious of the assumption that C
equilibrium constant expression gives the following result. The amount of H3O+ ion in water is so
for the approach taken to the calculation for acetic acid to
In other words, the acid must be weak enough that C
to calculate the equilibrium concentrations of H3O+,
can use successive approximations to solve the problem. illustrate this point, the next section will use both assumptions
ion concentration in these solutions are summarized below. here to check your answer to Practice Problem 1, Click
smaller than 1.0 x 10-13, the
acid in the previous section, we have to
= 1.1 x 10-2). ion in an aqueous solution gradually decreases and the pH of the
K a is a better measure of the strength of an acid than pH because adding water to an acid solution doesn't change its acid equilibrium constant, but does alter the H + ion concentration and pH. ion for each OH- ion when water dissociates, the
influence the H3O+ ion concentration in
Problems Involving Very Weak Acids. dissociation of water when KaCa
ignore the dissociation of water. concentration of the acid. We then solve this equation for the H3O+
it also must be strong enough that the H3O+
We have already confirmed the validity of the first
strength of the acid as reflected by the value of Ka,
pH of acetic acid solutions with the following
concentrations in pure water are the same. ion concentration at equilibrium also depends on the initial
Weak acids, as mentioned above, are those that are partially dissociated in water. H 3O+, OAc-, and HOAc
expression. HClO2(aq) + H2O(l)
water in our calculations.). Because we get one H3O+ ion for each OH-
the total. an equation, we start by assuming that we have a generic acid,
The following examples probe the
for this artificial sweetener. of water, we use the following equation to calculate the
expression. Rearranging this equation and taking the square root of both
ion from the dissociation of water is always equal to the amount
water, we get the second equation. is small compared with the initial concentration of the
must shift to the right to reach equilibrium. The value of Ka for this acid is
acetic acid in water. The first step, as always, involves building a representation
The total H3O+ ion concentration in an
ion concentration and take the square root of both sides. the quadratic formula, it is tempting to test the assumption that
Or we
We start this calculation by building a representation of what
The results of the previous two examples provide a basis for
up. of OH- ion from this reaction. ions from the dissociation of acetic acid. is small compared with the initial concentration of HOAc. When we can ignore the dissociation of
We can generate a more useful version of this equation by
Weak acids or bases can dissociate in an aqueous solution to achieve equilibrium. By convention, the symbol used to represent the initial
HA, that dissolves in water. The second assumption is hidden in the way the problem is set
dissociation of water. as the value of Ka
Calculate
solution increases as the solution becomes more dilute. 0.10 M solution of chlorous acid (Ka
dissociation of acetic acid is 0.0013 M. The OH-
The concentration of the HA molecules at equilibrium is equal
ion concentration in this solution is therefore 7.7 x 10-12
strong, C won't be small enough to be ignored. is therefore equal to the square root of Kw. only fails for dilute solutions of very weak acids. equation. constructing a model that allows us to predict when we can ignore
Two factors must be built into this model: (1) the
+ CN-(aq) Ka
remembering that we are trying to solve equilibrium problems for
Substituting the second equation into the first gives the
Equilibrium Problems Involving Strong Acids, Compounds that could be either Acids or Bases, Solving Equilibrium

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