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ISU CHE 141 - Aqueous Equilibrium

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CHEM 141 1st Edition Lecture 11Outline:- The Autoionization of Water- The pH Scale- Defining pOH- Interconverting pH and pOHAqueous EquilibriumThe Autoionization of Water- Water is amphiprotic- One water molecule, acting as an acid, donates a proton to another, which acts as a base and accepts the proton- The donor H2O forms its conjugate base (OH)- and accepts H2O forms its conjugate acid(H3O+):H2O(l)+H2O(l) reversible H3O+(aq)+OH-(aq)- This process is known as the autoionization of water- This process produces very small equal concentrations of H3O+ and OH- ions in pure water- We can write an equilibrium constant expression: Kw=[H3O+][OH-] or [H+]instead of [H3O+]- Kw is the equilibrium constant for water or ionic product- Heterogeneous equilibra: don’t include H2O(l)- H2O(l)+H2O(l) reversible H3O+(aq)+OH-(aq)o Experimentally, in pure water at 25 degrees C: [H3O+]=[OH-]=1.00x10-7Mo Substituting into Kw=[H3O+][OH-]o Kw=(1.00x10^-7M)(1.00x10-7M)o Kw=1.00x10^-14 which applies to all aqueous solutions 25 degree Co Kw is very tiny, confirming that only a small fraction of H2O molecules undergo autoionization- The reverse of autoionization is also occurring simultaneously: H3O+(aq)+OH-(aq) reversible H2O(l)+H2O(l)- The equilibrium constant for this reverse reaction is: Krev=1/Kw- Krev=1/1.00x10^-14=1.00x10^14- The value of Krev is very large so essentially the reverse of autoionization goes to completion These notes represent a detailed interpretation of the professor’s lecture. GradeBuddy is best used as a supplement to your own notes, not as a substitute.- We only need to consider the autoionization of H2O if acids/bases are extremely weak or very dilute- Kw=[H3O+][OH-]=1.00x10^-14- Inverse relationship between [H3O+] and [OH-]- As the value of one increases, the value of the other decreases so their product is always 1.00x10^-14- If [H3O+]>[OH-] then the solution is acidic- If [H3O+]<[OH-] then the solution is basic- If [H3O+]=[OH-]=1.00x10^-7M then the solution is neutralThe pH Scale- We can define how acidic a solution is by its concentration of H3O+, the greater [H3O+],the more acidic- Because [H3O+] in an aqueous solution is usually quite small, we usually express [H3O+] in terms of pH: pH=-log[H3O+]- Remember a logarithm is basically another way of writing indices- The base 10 logarithms of any number is the power to which 10 must be raised to equal the number- 10^y=x thus log x=y: 10^3=1000, therefore log 1000=3: [H3O+]=10^-pH- Therefore if y is negative (as it would be for small concentrations of [H+]), so is logx: 10^-2=0.01, therefore log 0.01=-2- Negative sign ensures most pH values are positive numbers between 0-14- Acidic solutions have [H3O+]>1.00x10^-7M which corresponds to pH<7.0- Basic solutions have [H3O+]<1.00x10^-7M which corresponds to pH>7.0- A solution with a pH of exactly 7.0 is neutral- Compare [H3O+] and pH: [H+] 1M(strong acid)----10^-7M(neutral)----10^-14(strong base)- pH 0(acidic)----7(neutral)----14(basic)- The logarithmic scale means that a change of one pH unit corresponds to a 10-fold change in [H+]:o A solution with pH=5.0 has 10 times the [H+] of a solution with pH=6.0o A solution with pH=12.0 has 1/10th the [H+] (or 10 times the [OH-]) as a solution with a pH of 11.0- Why is the pH=7 neutral? Kw=[H3O+][OH-]=1.00x10^-14- At 25 degree C, for pure water to be neither acidic nor basic: [H3O+]=[OH-]- Substitute [H3O+]=[OH-] into expression for Kw: Kw=[H3O+]^2=1.00x10^-14- Taking the square root of each side: [H3O+]=1.00x10^-7M: pH=-log[H3O+]:pH=-log(1.00x10^-7)=7.0Defining pOH- All aqueous solutions contain some H3O+ ions and some OH- ions from the autonizationof water- We can define a pH value for every aqueous solution form the negative logarithms of [H3O+]: pH=-log[H3O+]- In the same way, we can also calculate the negative logarithm of [OH-], which is called pOH: pOH=-log[OH-]: [OH-]=10^-pOHInterconverting pH and pOH- How does pOH relate to pH? Kw=[H3O+][OH-]=1.00x10^-14- Take –log of both sides- Log laws: logxy=logx+logy: -logKw=-(log[H3O+]+log[OH-])=-(-14.00): -logKw=-log[H3O+]-log[OH-])=14.00- Substitute in pH=-log[H3O+],pOH=-log[OH-], pKw=negative logarithm of Kw: pKw=pH+pOH=14.00- pKw is simply the logarithmic form of Kw- When we put a “p” in front of an equilibrium constant, it just means “-log”- Similarily, pKa is the logarithmic form of Ka (acid dissociation constant) and pKb is the logarithmic form of Kb (base protonation constant)- pH=-log[H3O+]:[H3O+]=10^-pH- pOH=-log[OH-]:[OH-]=10^-pOH- pKa=-logKa:Ka=10^-pKa- pKb=-logKb:Kb=10^-pKb- pKw=-logKw:Kw=10^-pKw- An aqueous solution at 25 degree C haspOH=13.92, calculate the pH: pKw=pH+pOH=14.00: pH=14.00-pOH:


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