What are examples of monoprotic, dip, and triprotic acids in chemistry?
The family of compounds known as monoprotic acids have been the focus of the acid equilibrium problems so far.When it acts as a Brnsted acid, each of these acids has a single H+ ion.monoprotic acids include hydrochloric acid and acetic acid.
Polyprotic acids can lose more than one H+ ion when they act as Brnsted acids.Diprotic acids have two acidic hydrogen atoms.phosphoric acid and citric acid have three.
The ease with which these acids lose the first and second protons is usually a large difference.Students assume that sulfuric acid loses its protons when it reacts with water because it is a strong acid.That is not a legitimate assumption.The Ka for the loss of the first protons is larger than 1.We assume that all the H2SO4 molecule lose the first protons to form HSO4-, or hydrogen sulfate, ion.
Only 10% of the H2SO4 molecule in a 1 M solution lose a second protons.
The values of Ka are given in the table.The large difference between the values of Ka for the sequential loss of protons by a polyprotic acid is important because it means we can assume that these acids dissociate one step at a time.
The assumption that polyprotic acids lose protons one step at a time is a consequence of the chemistry of a saturated solution of H2S in water.
rotten eggs have an unpleasant odor from hydrogen sulfide.It is an excellent source of the S2 ion and is used in introductory chemistry laboratories.H2S is a weak acid.The H2S molecule loses a protons in the first step of forming the hydrogen sulfide ion.
A small fraction of the H+ ion formed in this reaction is lost in a second step.
There are only four unknowns because the [H3O+] and [HS-] terms appear in both equations.The total H3O+ ion concentration from both steps must have the same value in both equations.The value of the balance between the first and second steps must be the same for both equations.
There are four equations that need to be solved.The Ka1 and Ka2 expressions are equations.We have to find two more equations or a pair of assumptions that can generate them.One assumption we can make is that the value of Ka1 is almost a million times larger than Ka2.
Only a small portion of the ion formed in the first step goes on to dissociation.Most of the H3O+ ion in this solution come from the dissociation of H2S, and the HS-ion formed in the reaction PSS remain in solution.The H3O+ and HS- ion concentrations can be assumed to be more or less equal.
We need one more equation and assumption.H2S is a weak acid.When the solution reaches equilibrium, we can assume that most of the H2S will be present.The equilibrium concentration of H2S can be assumed to be equal to the initial concentration.
Since there is always a unique solution to four equations in four unknowns, we are ready to calculate the H3O+, H2S, HS-, and S2 concentrations at equilibrium.A saturated solution of H2S in water has an initial concentration of 0.10 M.
We can use the equilibrium expression for the first step without having to worry about the second step.The expression for this acid is Ka1.
We are three-fourths of the way to our goal if our two assumptions are valid.We know the concentrations.
The values of three unknowns were obtained from the first equilibrium expression.
The following equation is given by substituting the known values of the H3O+ and HS- ion concentrations into this expression.
The equilibrium concentration of the H3O+ and HS- ion is the same as the value of Ka2 for this acid.
It's time to check our assumptions.The initial concentration is compared to the dissociation of H2S.Yes.The initial concentration of H2S is less than the 1.0 x 10-4 M ion concentrations obtained from this calculation.The following assumptions are valid.
Is the difference between the S2 and S3 concentrations large enough to allow us to assume that all of the H3O+ ion formed in the first step remain in solution?Yes.The calculation shows that the S2 ion concentration is much smaller than the HS concentration.Our other assumption is valid as well.
The following are the concentrations of the various components of this equilibrium.
We have used techniques with diprotic acids.The base values of Kb are the only challenge.
A solution that is initially 0.10 M in Na2CO3 will be used to calculate the concentrations.Ka1 is 4.5 x 10-7; Ka2 is 4.7 x 10.
The carbonate ion acts as a base toward water, picking up a pair of protons at a time to form the HCO3- ION, and then eventually carbonic acid, H2CO3.
Determining the carbonate ion values is the first step in solving the problem.The carbonate ion and the Ka expressions for carbonic acid are compared.
The expressions for Kb1 and Ka2 depend on the concentrations of HCO3- and CO32- ion.The expressions for Kb2 and Ka1 both depend on the HCO3- and H2CO3 concentrations.Ka2 and Kb2 can be calculated from Ka1.
The top and bottom of the Ka1 expression are combined to create the OH- ion concentration.
The first and second terms in this equation are inverse of the Kb2 expression.
We are ready to do the calculations.The strongest base in this solution is the CO32- ion, so we start with the Kb1 expression.
The difference between the carbonate ion and Kb2 is large enough to suggest that most of the OH- ions come from this step.
C is assumed to be small compared to the initial concentration of the carbonate ion because the value of Kb1 is small.The initial concentration of Na2CO3 will be roughly equal to the CO32- ion at equilibrium.
The equilibrium concentrations of OH-, HCO3-, and CO32- ion are calculated using the value of C.
The following result is given by substituting what we know about the OH- and HCO3- ion concentrations into this equation.
The H2CO3 concentration is equal to Kb2 for the carbonate ion according to this equation.
We can test the assumptions made in generating the results by summing them up.
Our assumptions are valid.The initial concentration of Na2CO3 is less than 5% of the reaction between the CO32- ion and water.Most of the HCO3- ion formed in the first step remains in solution.
Tripsrotic acids and bases can be worked on with our techniques.Let's use a 0.10 M H3PO4 solution to show this.
Let's assume that the first step is the most extensive reaction.
We assume that the difference between Ka1 and Ka2 is large enough that most of the H3O+ ion come from this first step.
The assumption that is small does not work in this problem.We can use the quadratic formula or successive approximations to solve the equation, so we don't need this assumption.We get the same answer either way.
The following equation is given by substituting what we know about the H3O+ and H2PO4- ion concentrations into this expression.
The HPO42- ion concentration is equal to Ka2 if our assumptions are correct.
The following equation is given by substituting what we know about the concentrations of H3O+ and HPO42 into this expression.
The only way to work this problem was to assume that the acid is one step at a time.The assumption is that all of the H3O+ ion come from the first step.Yes.Is it large enough to justify the assumption that all of the H2PO4- formed in the first step remains in solution?Yes.