CHE 141 1st Edition Lecture 14Outline- Calculating the pH of a Weak Base Solution- Polyprotic Acids- pH of Very Dilute Solutions- Strong Polyprotic AcidsCalculating the pH of a Weak Base Solution- We calculate the pH of a weak base solution in a similar way except we are solving for [OH-] not[H3O+]- Once we know [OH-], we can use the ionic product for water to calculate [H3O+]: Kw=[H3O+][OH-]=1.0x10^-14 at 25 degrees Celsius, pKw=pH+pOH=14 at 25 degrees Celsius- These expressions are equivalent p=-log- Once know [H3O+], we can calculate pH- The base protonation constant Kb of lidocaine (C14H21NONH) is 1.15x10^-8, calculate the pH of a 1.3 M solution of lidocaine at 25 degrees Celsius- Stratey:o Write the balanced equation for the reaction of the weak base with watero C14H21NONH is a weak base so need to write a base protonation constant expression and use an ICE table to calculate equilibrium concentrationso Calculate[H3O+]equilibrium using product of water(Kw) in order to then calculate pHo Kb=[C14H21NONH2+][OH-]/[C14H21NONH]=1.15x10^-8o C14H21NONH C14H21NONH2+ OH-o I 1.3 0 0o C -x +x +xo E 1.3-x x xo Kb=(x)(x)/(1.3-x)=1.15x10^-8o X=1.2x10^-4o Is weak base approximation valid? Check to see if x is less than 5% of initial concentration: percent ionization =[OH-]equilibrium/[B]initialx100%o (1.2x10^-4/1.3) x 100%=0.009%o less than or equal to 5% so x is small assumption is valido Substitute x into [OH-]=x, [OH-]=1.2x10^-4o Kw=[H3O+][OH-]=1.0x10^-14These 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.o H3O+=Kw/OH-=1.0x10^-14/1.2x10^-4=8.3x10^-11o pH=-logH+o pH=-log(8.3x10^-11)=10.1pH of Very Dilute Solutions- What is pH of 1.00x10^-8M HCl?o HCl is a strong acid so we assume complete ionizationo [H3O+]=[HCl]=1.00x10^-8o pH=-log(1.00x10^-8)=8o How can a solution of strong acid pH<7: it doesn’t because solution is very dilute,2 sources of H3O+ ions=ionization of the acid and auto ionization of water- How do we calculate H3O+ resulting from the autoionization of water?o Kw=[H3O+][OH-]=1.00x10^-14o Use x to represent [H3O+] and [OH-] resulting from autoionization o The [H3O+] term in Kw expression is the sum of [H3O+] from ionization of the acid and [H3O+] from autoionization of watero [H3O+]=(1.00x10^-8)+xo Substitute [H3O+]=(1.00x10^-8)+x and [OH-]=x into the Kw expressiono Plug that into the quadratic formula to get x=9.5x10^-8o Substitute that in: [H3O+]=(1.00x10^-8)+(9.5x10^-8)=1.05x10^-7o pH=-log[H3O+]o pH=-log(1.00x10^-6)=6o Autoionization of water is negligible in this caseo Rule: we only need to take into account the autoionization of water if the acid is diluted to a concentration <1.00x10^-6Polyprotic Acids- Monoprotic acids: acids which have only one ionizable hydrogen atom per molecule- We also need to consider polyprotic acids: acids which have more than one ionizable hydrogen- Adiprotic acid has two ionizable hydrogen atoms per molecule- A triprotic acid has three ionizable hydrogen atoms per molecule- Where there are multiple ionization steps, the value of Ka for each step is given a subscript- Ka1 denotes removal of the 1st H+ ion per molecule, Ka2 denotes removal of the 2nd H+ ion per molecule- Ka3 denotes removal of the 3rd H+ ion per molecule- For all polyprotic acids Ka1>Ka2- Consider electrostatic attractions:o More difficult to remove a second H+ ion(positive charge) from a negatively charged anion (Ka2)o To remove a third H+ ion from an ion with a -2 charge is even more difficult (Ka3)o Typically only the first ionizable hydrogen (Ka1) affects the strength (pH) of a polyprotic acido Rule: If the difference between Ka1 and Ka2 is greater than or equal to 10^3, thenwe will only need to consider Ka1- What is the pH of a .150M solution of the diprotic acid, sulfurous acid (H2SO3), given the following ionization equilibria?o H2SO3+H2O reversible HSO3-+H3O+: Ka1=1.7x10^-2o HSO3-+H2O reversible SO32-+H3O+:Ka2=6.2x10^-8o Ka1>>Ka2 by a factor greater than or equal to 10^3 thus the pH is controlled by the first ionization equilibriumo Rule: treat weak polyprotic acids with Ka1>>Ka2 by a factor greater than or equalto 10^3 like a normal calculation of the pH of a weak acido Make and ICE table then plug the equilibriums into the Ka1 equation o Solve for x to get x=0.050o Then do the percent ionization=(0.050/0.150)x100%=33.7%o Use quadratic formula to get x=0.0427o Then do pH=-log(0.0427)=1.37Strong Polyprotic Acids- We only know one strong polyprotic acid, H2SO4:- H2SO4HSO4-+H+ : Ka1>>1- HSO4- reversible SO42-+H+: Ka2=1.2x10^-2- If [H2SO4] is greater than or equal to 1.0M only the first dissociation makes significant contribution to [H3O+] and thus pH- If [H2SO4]<1M then we need to consider the contribution of the second ionization to [H3O+] to calculate pH- Calculate the pH of a 0.050M solution of sulfuric acid: H2SO4HSO4+H+:Ka1>>1o HSO4 reversible SO42-+H+: Ka2=1.2x10^-2o The total [H3O+] will be: [H3O+] from ionization of H2SO4 (100%) ionization +[H3O+] from partial ionization of HSO4-o For the second ionization ICE table: [HSO4-]initial=[H2SO4]initial from the first ionization (from stoichiometry) as H2SO4 fully ionizes to H3O+ and HSO4-, [H3O+]initial=[H3O+] from first ionization=[H2SO4] (from stoichiometry), SO42-initial=0o {H3O+] after first ionization=[H2SO4]=0.050M thus [H3O+] initial for second ionization=0.050Mo [HSO4-]initial for second ionization=0.050Mo Set up ICE table, plug into Ka2 and solve quadratic get x=0.00851, then solve for pH and get it to