Salts and Buffers Ch17 4.17.2

Acid – Base Properties of Salt Salt is another name for ionic compound When a salt dissolves in water, it breaks into its ions Under certain situations, these ions can behave like acids and bases, affecting the pH of solutions!

3 ways a salt can affect pH A salt can dissociate to form: A neutral soln (pH = 7) An acidic soln (pH < 7) A basic soln. (pH > 7) How do we predict how a salt will behave? We look at the cation and anion of the dissociated salt.

Salts that Produce Neutral Solutions Contain conjugate bases of strong acids Contain cations from strong bases Strong acids Conjugate Base Example of salt HCl Cl- NaCl HBr Br- NaBr HI I- KI HClO3 ClO3- KClO3 Strong Base Cation Example of salt NaOH Na+ NaCl CsOH Cs+ CsI Ca(OH)2 Ca+ Ca(ClO3)2 Sr(OH)2 Sr2+ SrSO4

Salts that produce neutral solutions NaOH Na+ + OH- HCl H+ + Cl- Neutralization reaction

Neutral Solutions- WHY? So why does a salt containing the conjugate base of a strong acid and the conjugate acid of a strong base produce a neutral solution? HCl NaOH Strong acid = Strong base = b/c so weak neither will be gaining H+ or give up H+ so don’t affect pH as pH depends on [H+]

Salts that produce basic solutions Salts derived from a strong base and a weak acid NaC2H3O2 Na+ (aq) + C2H3O2-(aq) C2H3O2-(aq) + H2O HC2H3O2 + OH- Why? b/c conjugate base of the weak acid is a strong base It deprotonates water thus increasing [OH-], increasing basicity H2O

So, the weak salt’s ANION must be the conjugate base of a weak acid The salts CATION must be the cation of strong base salt Cation and Strong base Anion and weak acid Sr(CN)2 Sr 2+ Sr(OH)2 CN- HCN KF K+ KOH F- HF NaNO2 Na+ NaOH NO2 - HNO2

Why is this useful Why would we need to know the type of solution that a salt produces/ Well, we are interested in pH for all kinds of important reactions, like controlling the pH of our blood and water sources. These things contain salts, which affect this pH. So how do we do pH calculations for solutions containing a salt??

Calculate the pH of a 0.30M NaF solution. The Ka for HF =7.2x10-4 NaF Na+ + F- (from SB) (from WA) step 1: identify major species Step 2: write dominate reaction

Calculate the pH of a 0.30M NaF solution. The Ka for HF =7.2x10-4 Step 3: calculate Kb, using Ka and Kw Ka x Kb = Kw Acid Conj base Kb=Kw/Ka (for HF) Kb =

Calculate the pH of a 0.30M NaF solution. The Ka for HF =7.2x10-4 Step 4 make icebox F- + H2O (l) HF + OH-