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16-11 Lewis Acids and Bases -- Part 3
How much solution is produced by mixing water with 2.50 L of solution?
24.80 mL of 0.248 M HNO3 and 15.40 mL of 0.394 M were used to calculate Ka.
Explain what the 17-4 Neutralization Reactions and common-ion effect is and how it relates to Titration Curves Le Chatelier's principle.
Explain how a buffer solution is able to resist change.
Discuss the method by which a pH indicator can be used.
There are difficulties in calculating the pH of a solution containing a salt of a polyprotic acid.
The step-by-step process of performing acid-base equilibrium calculations is summarized.
Richard Megna/Fundamental Photographs NaOH(aq) is slowly added to the solution.
The indicator color changes as the pH goes from 8.0 to 10.0.
When the solution turns a pink, the neutralization point is reached.
The selection of indicators for acid-base titrations is one of the topics considered in this chapter.
A small amount of atmospheric CO2(g) can be found in acid rain.
The amount is enough to lower the pH of the rain.
When acid-forming air pollutants, such as SO2, SO3 and NO2, are dissolved in rain, it becomes even more acidic.
A chemist would say that water doesn't have a "buffer capacity" because its pH changes quickly when small quantities of acids or bases are dissolved in it.
buffer solutions can resist a change in pH when acids or bases are added to them.
There are additional aspects of acid base equilibria.
The buffer system that maintains the constant pH of blood is perhaps the most important buffer system to humans.
Acid-base titrations is one of the topics that we will explore.
We want to calculate how pH changes during a titration.
This information can be used to determine which acid-base titrations work well and which don't.
The calculations in this chapter are extensions of those in Chapter 16.
Buffers are used to determine the concentrations of the species present at equilib.
A solution of a weak acid or weak of the common-ion effect in base initially contains a second source of one of the ion produced in the ion weak acids and weak bases.
There are some consequences to the presence of a common ion.
We can write separate equations for the ionizations of the acids.
The common ion, H3O+, appears in the equilibrium constant expressions for both reactions.
We will look at the extent to which the other acid affects the acid's ionization.
In order to do this, we will write the ionized constant for a general acid, HA, in terms of the degree of ionization, an Loon a, which we introduced in Section 16-3.
Since a represents the fraction of HA that exists as A- at equilibrium, then (1 - a) is the fraction that exists as HA.
We can use this expression to calculate the degree of ionization of an indicator.
If we know the pH of the solution, we can compare this photo acid.
The pH is equal to 1.0.
We can see that HCl is completely ionized in this solution.
The presence of a strong acid suppressed the ionization of a weak acid.
In a solution of 0.
100 M, determine the same quantities.
We must determine the species in a weak acid solution and investigate the effects of adding a strong acid.
The acids have the same ion.
The key to solving part (b) is knowing that HCl, a strong acid, ionizes completely, regardless of whether or not any other acids are present in the solution.
The H3O+ is produced by the ionization of HCl.
We can see that x is 10.
The equation (17.3) can be used for the mixture of acids.
Determine the concentrations in a solution of 100 M HCl and 500 M HF.
The volume of solution will remain at 1.00 L after the 12 M HCl is removed.
The presence of a strong acid suppresses the ionization of a weak acid.
The presence or addition of a strong base significantly suppressed the ionization of a weak base.
The statements can be justified by applying Le Chatelier's principle.
The solution of a weak acid, HA, has reached equilibrium.
The effect of adding strong acid is shown.
The HA is not being ionized.
The effect of adding a strong base can be described as similar to the effect of adding a weak base.
The equilibrium shifts when a strong base supplies OH2.
The calculated pH is 4.74.
An example of a situation involving the common ion effect is when dealing with a solution containing a weak acid (HA) and a salt of meters used.
The anion of the and their accuracy is probably salt, which is produced by the ion of the acid and less than that.
The CH COOH is not being ionized.
If you want to solve a common-ion problem, you should assume that the weak acid and its salt have been placed in solution.
Ionizing occurs until equilibrium is reached.
3H3O+4 and 3CH 3COO 4 are calculated in a solution of 0.
This example is very similar to example 17; however, in this case we will be adding a salt of a weak acid and observing the shift in equilibrium.
The NaCH COOs should be solved completely.
This is a valid assumption.
The salt that was added reduced the ionization of CH3CO OH.
In this example, calculate 3H3O+4 and 3HCOO-4 in a solution of 100 M HCOOH and 0.150 M NaHCOO.
The volume should remain at 1.00 L.
The weak acid-anion situation is similar to the common-ion effect of a salt.
The indicator is blue if the pH is 7 and 10.
The key results of this section can be summarized.
The pH drops below 10 due to the suppression of the ionization of the weak acid, HA.
The solution contains equilibrium amounts of both CH COOH and its conjugate base.
A buffer solution is able to resist changes in pH because it contains components capable of neutralizing other acids or bases but not each other.
Buffer solutions are very useful.
Identifying the active components of a buffer solution is the first step in our study.
We will discuss how these components give a buffer the ability to resist.
We will develop an equation that shows the relationship between the concentrations of the two active components and the pH of the buffer solution.
The practical matter of preparing a buffer solution with a specified pH will be considered.
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