Example:
Salt: NaCl
Parents: NaOH/HCl
pH: neutral (a strong acid and a strong base)
Salt: NaF
Parents: NaOH/HF
pH: basic (strong base, weak acid)
Salt: NH4Br
Parents: NH3/HBr
pH: acidic (strong acid, weak base)
Salt: NH4NO2
Parents: NH3/HNO2
pH: must find Ka, Kb and compare
But what do we need to find the Ka and Kb for? When dealing with a salt, follow solubility rules. All salts of ammonium are soluble, so the salt will break apart in water:
NH4NO2 (aq) → NH4+ (aq) + NO2- (aq)
These ions will go on to react with water, where one will produce OH- (giving a basic property) and one will produce H3O+ (to produce an acidic quality). Whether the solution itself is acidic or basic depends on whether more OH- is produced or more H3O+ is produced. In other words, we can find whether the solution is acidic or basic by seeing which is larger, the Ka or the Kb:
NH4+ (aq) + H2O (l) ⇌ H3O+ (aq) + NH3 (aq)
NO2- (aq) + H20 (l) ⇌ OH– (aq) + HNO2 (aq)
Because the reaction with NH4+ produces H3O+, we know we need to find the Ka for this reaction. The table doesn't list the Ka for NH4+. How do we find it?
The product of Ka and Kb for conjugate acid-base pairs equals Kw.
Ka x Kb = Kw
So to find the Ka for NH4+, we find the Kb of the conjugate base, and plug these numbers into the above equation to solve for Ka.
KaNH4+ = Kw / KbNH3 = (1 x 10-14) / (1.8 x 10-5) = 5.56 x 10-6
Because the reaction with NO2- produces OH-, we know we need to find the Kb for this reaction. The table doesn't list the Kb for NO2-. How do we find it?
Again, find the Ka for the conjugate acid, and plug these numbers into the above equation and solve for Kb.
KbNO2- = Kw / KaHNO2= (1 x 10-14) / (4.5 x 10-4) = 2.22 x 10-11
Here, Ka is greater than Kb, so the solution will be acidic.
Example:
What is the pH for 0.250 M KClO?
KClO is a salt. According to solubility rules, it is soluble:
KClO (aq) → K+ (aq) + ClO-(aq)
These ions will go on to react with water, where one will produce OH- (giving a basic property) and one will produce H3O+ (to produce an acidic quality).
K+ (aq) +H2O (l) ⇌ KOH (aq) + H3O+ (aq)
ClO- (aq) + H2O (l) ⇌ HClO (aq) OH- (aq)
But there's a catch. KOH is a strong base, and we know that strong bases dissociate (ionize) 100% in water. So really,
K+ (aq) + H2O (l) → No Reaction
We also know that we had 0.250 M KClO to begin with. The KClO dissociated into K+ and ClO- where everything is 1:1, so it dissociates to form 0.250 M of K+ (that produces no reaction) and 0.250 M ClO-, which will go on to react with water. This is a little more obvious when we build a table:
Now that we know we have 0.250M ClO-, we can build our table for the reaction with water:
ClO- (aq) + H2O (l) ⇌ HClO (aq) OH- (aq)
Hint: we know we need to find Kb because OH- is a product
Kb = (x2) / (0.250 – x) → approximation method holds
How do we find the Kb for ClO-?
KbClO- = Kw / KaHClO = (1 x 10-14) / (3.5 x 10-8) = 2.89 x 10-7
So: 2.89 x 10-7 = (x2) / 0.250
x = 2.69 x 10-4 = [OH-]
pOH = -log(2.69 x 10-4) = 3.57
pH = 14 – 3.57 = 10.4
Example:
What is the pH for 0.135M NH4Cl?
NH4Cl is a salt, which ions will it form when it dissociates?
NH4+ and Cl-
NH4Cl → NH4+ + Cl-
These ions with go on to react with water. We know Cl- will react to form HCl, which is a strong acid, and as such it will dissociate 100% in water, so overall, there is no reaction. So what happens to the NH4+? It is the conjugate acid of a weak base, so:
NH4+ (aq) + H2O (l) ⇌ NH3 (aq) + H3O+ (aq)
Again, it is easy to see that since we start out with 0.135 M of NH4Cl, we will get 0.135 M of each of the ions (1:1), so when we build a table for the reaction of the ammonium ion with water, we know to start with 0.135 M.
Hint: we know we need to find Ka because H3O+ is a product
Ka = (x2) / (0.135 – x) → approximation holds
How do we find the Ka for NH4+?
KaNH4+ = Kw / KbNH3 = (1 x 10-14) / (1.8 x 10-5) = 5.56 x 10-10
So: 5.56 x 10-10 = (x2) / (0.135)
x = 8.66 x 10-6 = [H3O+]
pH = -log(8.66 x 10-6) = 5.06
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