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16-11 Lewis Acids and Bases -- Part 4
The term "appreciable" means that a buffer solution contains either a weak acid (HA) and its conjugate base (A-), or a weak acid (B) and its conjugate acid (BH+).
The solution will be able to neutralize either an added acid or an added base with both components present.
We must add two components to the solution to get enough active components.
A- is never produced by the ionization of a weak acid.
The weak base of A- never produces an adequate amount of HA.
A buffer solution is CH COOH and 0.
100 M Na CH COO.
The H O is 1.8 and the pH is 4.74.
A buffer solution has the ability to add strong acid or strong base to it's composition.
If the acid from the buffer or base is strong, there is a weak acid completion.
We have a solution with a ratio of 3CH3COOH4 L 3CH3COO-4 and a solution with a ratio of 3CH3COOH4>3CH3COO-4.
The conjugate base of acetic acid acts as a "sink" when strong acid is added.
The ratio is kept constant so there is no change in the pH.
acetic acid acts as a proton donor when strong base is added, keeping the ratio approximately constant and minimizing the change in pH.
The pH is very close to the original value.
At the beginning of the section, we pointed out that a buffer contains components that can counteract an added acid or base but not each other.
It would have been better to say that the two components coexist in the buffer.
Predicting whether a solution is a buffer solution is part of the CH3 exam.
To show that a solution has buffer properties, first identify a component in the solution that neutralizes acids and a component that neutralizes bases.
NH3 and NH + 4 are included in this example.
The equations below show how 4 counteracts strong base.
Not all NH3-NH4Cl buffer solutions will be effective.
A buffer solution can be created by a mixture of a strong acid and the salt of a weak acid.
A mixture of NH3 and HCl can result in a buffer solution.
This example is very similar to another example.
We need to calculate the molarity of the acetate ion before we solve the equilibrium part.
The molarity of CH 3COO in 500.0 mL of solution is calculated.
This assumption will be valid.
When acetic acid and acetate ion are present in equal concentrations, we have seen that pH is 4.74.
The solution should be less acidic.
A pH of 4.80 is a reasonable answer.
A handbook states that to prepare 100.0 mL of a particular buffer solution, mix 63.0% of 0.200 M CH3CO OH with 33.0% of 0.200 M Na CH3COO.
This equation is used by biologists.
Let's consider a mixture of a hypothetical weak acid, HA, and its salt, NaA, to derive this variation of the ionization constant expression.
rearrange the equation to solve for pH.
When the eral equation (17.7), the Henderson-Hasselbalch equation, is written, A- is the conjugate base of the weak acid HA.
3conjugate base4 is 3acid4 and the pH is pKa.
It is important to avoid pitfalls of using limitations on the equation's validity.
When the equation is not valid, we will see.
If the value of H3O+ is very small compared to the values of acid and base, then the use of concentrations in the equation is justified.
We need a buffer solution with a pH of 5.
The equation suggests a conjugate base and acid.
One way to find a weak acid is to find a solution with equal molarities of the acid and its salt.
Although it is simple in concept, it is not practical.
We are not likely to find a readily available water-soluble weak acid.
To get a pH of 5.09.
The solution volume should remain constant at 0.300 L.
An equilibrium concentration is the number of concentration terms in a Ka expression.
The 3H3O+4 is the equilibrium concentration.
We will assume that the equilibrium concentration is the same as the initial concentration.
The equilibrium concentration is what we calculate with the Ka expression, and we will assume that it is the same as the stoichiometric concentration.
If the conditions are met, these assumptions work well.
We have finished the calculation of the mass of sodium acetate.
The answer is checked by putting the acetic acid concentrations, along with the pKa of acetic acid, into the equation.
Adding an appropriate amount of strong base to 0.300 L of 0.25 M CH3COOH1aq2 is one approach.
A weak acid or a weak base can be used to prepare a buffer solution for an experiment.
This is a method of getting a buffer solution.
Other methods can be useful.
A mixture of an amine and its conjugate acid is a buffer solution.
Weak acids and amines can be prepared in similar ways.
The new concentrations of weak acid and its salt can be used to calculate the pH of the buffer solution.
The problem is solved in two steps.
The equilibrium constant expression is solved for 3H3O+4 by substituting the new concentrations into it.
There are two parts of the calculation.
The conjugate acid-base pair BH+>B can be applied to this scheme.
We complete the same calculations in parts a and b.
Adding a strong acid or a strong base to the buffer solution is something we should be aware of.
To investigate this effect, we have to make a calculation.
The neutralization of the base or acid components of the buffer is taken into account in the calculation.
The value of the original buffer's pH is what we must keep in mind when judging the effect of acid or base on it.
The initial pH of the buffer can be obtained by substituting the initial concentrations into an equation.
Assume that the neutralization goes to completion.
This is a limiting reactant calculation that is simpler than many of the ones in Chapter 4.
The new equilibrium concentrations can be used to calculate the pH.
CH 3COO is converted to OH- in the added OH-.
The last line of the table shows the calculation of the new concentrations.
The same type of calculation is used, but with slightly different concentrations.
The water's pH would have changed by more than 5 units.
The results are reasonable.
A 1.00 L volume of buffer is made with concentrations of 0.350 M NaHCOO and 0.550 M HCOOH.
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