The mountain equilibrium problems questioned so far have concentrated on a family members of compoundsknown together **monoprotic acids**. Every of these acids has a solitary H+ ion, orproton, it can donate when it acts as a Brnsted acid. Hydrochloric mountain (HCl), aceticacid (CH3CO2H or HOAc), nitric mountain (HNO3), and benzoicacid (C6H5CO2H) space all monoprotic acids.

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Several essential acids have the right to be classified together **polyprotic acids**, which can losemore 보다 one H+ ion when they act together Brnsted acids. **Diprotic acids**,such as sulfuric mountain (H2SO4), carbonic acid (H2CO3),hydrogen sulfide (H2S), chromic acid (H2CrO4), and oxalicacid (H2C2O4) have actually two acidic hydrogen atoms. **Triproticacids**, such together phosphoric acid (H3PO4) and citric mountain (C6H8O7),have three.

There is typically a huge difference in the ease with which this acids lose the firstand second (or second and third) protons. As soon as sulfuric acid is classified as a strongacid, students frequently assume the it loser both that its protons as soon as it reacts through water.That isn"t a legitimate assumption. Sulfuric mountain is a strong acid since Kafor the loss of the very first proton is much larger than 1. We because of this assume thatessentially every the H2SO4 molecule in one aqueous solution shed thefirst proton to form the HSO4-, or hydrogen sulfate, ion.

H2SO4(aq) + H2O(l) " width="17" height="9" sgi_fullpath="/disk2/princetoneclub.orgistry/genprincetoneclub.org/public_html/topicreview/bp/ch17/graphics/rarrow.gif"> H3O+(aq) + HSO4-(aq) | Ka1 = 1 x 103 |

But Ka because that the lose of the 2nd proton is only 10-2 andonly 10% the the H2SO4 molecule in a 1 M solution lose asecond proton.

HSO4-(aq) + H2O(l) | Ka2 = 1.2 x 10-2 |

H2SO4 just loses both H+ ions as soon as it reacts with abase, such together ammonia.

The table listed below gives worths of Ka because that some typical polyprotic acids.The huge difference in between the values of Ka because that the sequential ns ofprotons by a polyprotic acid is important because it method we have the right to assume that these acidsdissociate one step at a timeanassumption recognized as **stepwise dissociation**.

Acid-Dissociation Equilibrium Constants for usual Polyprotic Acids

Acid | Ka1 | Ka2 | Ka3 | |||

sulfuric acid (H2SO4) | 1.0 x 103 | 1.2 x 10-2 | ||||

chromic acid (H2CrO4) | 9.6 | 3.2 x 10-7 | ||||

oxalic mountain (H2C2O4) | 5.4 x 10-2 | 5.4 x 10-5 | ||||

sulfurous mountain (H2SO3) | 1.7 x 10-2 | 6.4 x 10-8 | ||||

phosphoric mountain (H3PO4) | 7.1 x 10-3 | 6.3 x 10-8 | 4.2 x 10-13 | |||

glycine (C2H6NO2) | 4.5 x 10-3 | 2.5 x 10-10 | ||||

citric mountain (C6H8O7) | 7.5 x 10-4 | 1.7 x 10-5 | 4.0 x 10-7 | |||

carbonic mountain (H2CO3) | 4.5 x 10-7 | 4.7 x 10-11 | ||||

hydrogen sulfide (H2S) | 1.0 x 10-7 | 1.3 x 10-13 |

Let"s look in ~ the repercussion of the presumption that polyprotic acids lose protons onestep at a time by analyzing the princetoneclub.orgistry of a saturated equipment of H2S inwater.

Hydrogen sulfide is the foul-smelling gas that offers rotten eggs their unpleasant odor.It is great source that the S2- ion, however, and also is because of this commonlyused in introduce princetoneclub.orgistry laboratories. H2S is a weak mountain thatdissociates in steps. Several of the H2S molecules lose a proton in the very first stepto kind the HS-, or hydrogen sulfide, ion.

First step: | H2S(aq) + H2O(l) |

A small fraction of the HS- ions developed in this reaction then go on to loseanother H+ ion in a second step.

Second step: | HS-(aq) + H2O(l) |

Since there room two procedures in this reaction, we have the right to write 2 equilibrium constantexpressions.

Although each of this equations contains three terms, there are only 4 unknowns

Four equations are required to fix for 4 unknowns. We already have 2 equations:the Ka1 and also Ka2 expressions. We are going to have actually tofind one of two people two more equations or a pair of assumptions that have the right to generate 2 equations.We have the right to base one presumption on the reality that the worth of Ka1 because that thisacid is virtually a million times larger than the worth of Ka2.

Ka1 >> Ka2

This way that only a small portion of the HS- ions created in the firststep walk on to dissociate in the second step. If this is true, most of the H3O+ions in this systems come indigenous the dissociation the H2S, and most the the HS-ions developed in this reaction PSS stay in solution. Together a result, we deserve to assume the the H3O+and HS- ion concentrations are much more or less equal.

First assumption: |

We require one much more equation, and therefore one much more assumption. Keep in mind that H2Sis a weak acid (Ka1 = 1.0 x 10-7, Ka2 = 1.3x 10-13). Thus, we can assume that most of the H2S the dissolves inwater will certainly still be existing when the systems reaches equilibrium. In various other words, we canassume that the equilibrium concentration the H2S is approximately equal come theinitial concentration.

Second assumption: |

We currently have 4 equations in four unknowns.

Since over there is constantly a distinctive solution to 4 equations in 4 unknowns, we space nowready to calculate the H3O+, H2S, HS-, and S2-concentrations at equilibrium in a saturated systems of H2S in water. All weneed to understand is that a saturated solution of H2S in water has actually an initialconcentration of about 0.10 M.

Because Ka1 is so much larger than Ka2 because that thisacid, we can work through the equilibrium expression for the very first step without worryingabout the 2nd step for the moment. We thus start v the expression because that Ka1for this acid.

We climate invoke one of our assumptions.

Substituting this approximation right into the Ka1 expression offers thefollowing equation.

We climate invoke the various other assumption.

*C*

Substituting this approximation into the Ka1 expression gives thefollowing result.

We currently solve this almost right equation because that *C.*

*C*1.0 x 10-4

If our two presumptions are valid, we space three-fourths of the means to our goal. We knowthe H2S, H3O+, and HS- concentrations.

Having extract the values of three unknowns indigenous the first equilibrium expression, weturn come the second equilibrium expression.

Substituting the well-known values that the H3O+ and HS- ionconcentrations right into this expression gives the complying with equation.

Because the equilibrium concentrations of the H3O+ and HS-ions are an ext or much less the same, the S2- ion concentration in ~ equilibrium isapproximately equal to the value of Ka2 for this acid.

It is now time to examine our assumptions. Is the dissociation that H2S smallcompared v the early concentration? Yes. The HS- and H3O+ion concentrations acquired from this calculation space 1.0 x 10-4 M,which is 0.1% of the initial concentration that H2S. The following assumption istherefore valid.

Is the difference between the S2- and HS- ion concentrationslarge enough to enable us come assume that essentially all of the H3O+ions in ~ equilibrium are created in the very first step and that essentially all of the HS-ions formed in this step remain in solution? Yes. The S2- ion concentrationobtained from this calculation is 109 times smaller sized than the HS- ionconcentration. Thus, our other presumption is additionally valid.

We can therefore summarize the concentrations of the various contents of thisequilibrium as follows.

**Diprotic Bases **

The methods we have actually used with diprotic acids have the right to be expanded to diprotic bases. Theonly challenge is calculating the values of Kb for the base.

Example: Let"s calculation the H2CO3, HCO3-,CO32-, and also OH- concentrations at equilibrium in asolution the is initially 0.10 M in Na2CO3. (H2CO3:Ka1 = 4.5 x 10-7; Ka2 = 4.7 x 10-11)

Because the is a salt, salt carbonate dissociates into its ions when it disappear inwater.

H2O | |||||

Na2CO3(aq) | " width="17" height="9" sgi_fullpath="/disk2/princetoneclub.orgistry/genprincetoneclub.org/public_html/topicreview/bp/ch17/graphics/rarrow.gif"> | 2 Na+(aq) | + | CO32-(aq) |

The lead carbonate ion then acts together a base towards water, choose up a pair of proton (oneat a time) to type the bicarbonate ion, HCO3- ion, and theneventually carbonic acid, H2CO3.

CO32-(aq) + H2O(l) | Kb1 = ? | ||

HCO3-(aq) + H2O(l) | Kb2 = ? |

The very first step in fixing this trouble involves identify the values of Kb1and Kb2 because that the carbonate ion. We begin by to compare the Kbexpressions for the carbonate ion through the Ka expressions because that carbonicacid.

The expressions for Kb1 and Ka2 have actually something incommontheyboth depend on the concentrations of the HCO3- and CO32-ions. The expressions for Kb2 and Ka1 also havesomething in commontheyboth depend on the HCO3- and also H2CO3concentrations. We can thus calculate Kb1 indigenous Ka2and Kb2 indigenous Ka1.

We begin by multiply the top and also bottom that the Ka1 expression bythe OH- ion concentration to introduce the

We then team terms in this equation together follows.

The very first term in this equation is the inverse of the Kb2 expression,and the 2nd term is the Kw expression.

Rearranging this equation offers the following result.

Ka1Kb2 = Kw

Similarly, we can multiply the top and also bottom of the Ka2 expressionby the OH-ion concentration.

Collecting terms gives the following equation.

The very first term in this equation is the station of Kb1, and the secondterm is Kw.

This equation can thus be rearranged together follows.

Ka2Kb1 = Kw

We can now calculation the values of Kb1 and also Kb2 forthe carbonate ion.

We are ultimately ready to perform the calculations. We begin with the Kb1expression due to the fact that the CO32- ion is the the strongest base in thissolution and therefore the best resource of the OH- ion.

The difference between Kb1 and also Kb2 because that thecarbonate ion is huge enough to indicate that many of the OH- ion come fromthis step and most that the HCO3- ions created in this reaction remainin solution.

*C*

The value of Kb1 is little enough to assume that *C*is little compared with the initial concentration of the lead carbonate ion. If this is true,the concentration the the CO32- ion at equilibrium will be roughlyequal to the initial concentration of Na2CO3.

Substituting this info into the Kb1 expression offers thefollowing result.

This approximate equation can now be solved for *C*.

*C*0.0046 M

We then use this value of *C*to calculation the equilibrium concentration of the OH-, HCO3-,and CO32- ions.

*C*0.095M

*C* 0.0046 M

We now turn come the Kb2 expression.

Substituting what us know around the OH- and HCO3- ionconcentrations into this equation provides the following result.

According to this equation, the H2CO3 concentration atequilibrium is approximately equal to Kb2 because that the lead carbonate ion.

Summarizing the outcomes of our calculations enables us to test the assumptions madegenerating this results.

All of our assumptions are valid. The extent of the reaction between the CO32-ion and also water to offer the HCO3- ion is much less than 5% of the initialconcentration that Na2CO3. Furthermore, most of the OH- ioncomes native the very first step, and also most of the HCO3- ion created in thisstep remains in solution.

**Triprotic mountain **

Our approaches for working diprotic mountain or diprotic base equilibrium difficulties can beapplied to triprotic acids and bases as well. To illustrate this, let"s calculate the H3O+,H3PO4, H2PO4-, HPO42-,and PO43- concentration at equilibrium in a 0.10 M H3PO4solution, because that which Ka1 = 7.1 x 10-3, Ka2 =6.3 x 10-8, and Ka3 = 4.2 x 10-13.

Let"s assume the this acid dissociates through steps and also analyze the first stepthemost extensive reaction.

We currently assume that the difference between Ka1 and also Ka2is large enough that many of the H3O+ ions come indigenous this first stepand most of the H2PO4- ions created in this step continue to be insolution.

*C*

Substituting this assumption into the Ka1 expression offers thefollowing equation.

The presumption that is little compared v the early concentration that the acid stops working in this problem. Yet wedon"t really need this assumption because we have the right to use the quadratic formula or successiveapproximations to solve the equation. One of two people way, we achieve the exact same answer.

*C*= 0.023 M

We deserve to then usage this value of *C*to achieve the complying with information.

*C*0.077 M

We currently turn to the second strongest mountain in this solution.

Substituting what we know around the H3O+ and also H2PO4-ion concentrations right into this expression offers the complying with equation.

If our assumptions so much are correct, the HPO42- ionconcentration in this systems is same to Ka2.

We have actually only one much more equation, the equilibrium expression for the weakest acid in thesolution.

Substituting what we know around the concentrations of the H3O+and HPO42- ions right into this expression gives the complying with equation.

This equation can be addressed for the phosphate ion concentration in ~ equilibrium.

Summarizing the outcomes of the calculations helps us inspect the presumptions made alongthe way.

The only approximation supplied in functioning this trouble was the presumption that the aciddissociates one step at a time. Is the difference between the concentrations of the H2PO4-and HPO42- ions huge enough to justification the presumption thatessentially every one of the H3O+ ions come native the an initial step? Yes. Isit large enough to justification the assumption that essentially all of the H2PO4-formed in the very first step remains in solution? Yes.

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You might never encounter an example of a polyprotic mountain for whichthe difference in between successive values of Ka are too tiny to enable usto i think stepwise dissociation. This assumption works also when we can expect that tofail.