pH of salt solutions

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Maldonado3K
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Joined: Fri Sep 29, 2017 7:07 am

pH of salt solutions

Postby Maldonado3K » Sun Jun 10, 2018 11:48 pm

Does anyone have any basic steps or simple explanation to determining the pH of a salt solution? Basic example?

Florence-1F
Posts: 11
Joined: Wed Nov 15, 2017 3:03 am

Re: pH of salt solutions

Postby Florence-1F » Mon Jun 11, 2018 1:56 pm

Find the concentration of the salt solution, whether the salt is an acidic, basic, or neutral salt, the equation for the interaction of the ion with the water, the equilibrium expression for this interaction and the Ka or Kb value.

Example: Calculate the pH of a 0.500 M solution of KCN. Ka for HCN is 5.8 x 10-10.

First, write the equation for the dissolving process, and examine each ion formed to determine whether the salt is an acidic, basic, or neutral salt.
KCN(s) --> K+(aq) + CN-(aq)
K+ is a neutral ion and CN- is a basic ion. KCN is a basic salt.

Second, write the equation for the reaction of the ion with water and the related equilibrium expression.
CN-(aq) + H2O(l) --> HCN(aq) + OH-(aq)
Kb = [HCN][OH-]
[CN-]

Third, use the given Ka for HCN to find the value of Kb for CN-.
(5.8 x 10-10)(Kb) = 1 x 10-14
Kb = 1.7 x 10-5
Make an "ICE" chart to aid in the solution. Let "x" represent the amount of CN- that interacts with the water.

Initial Concentration ---------- Change in Concentration --------------- Equilibrium Concentration
0.500 M : CN-(aq) ----------- -x ------------------ 0.500 M - x
0 M : HCN(aq) -------------- +x ------------------- x
0 M : OH-(aq) -------------- +x --------------------------- x


Subsititute equilibrium values and the value for Kb to solve for x.
1.7 x 10-5 = (x)(x)/(0.500 - x)
We will make the assumption that since Kb is so small that the value for x will be very small as well, thus the term (0.500 - x) is equal to (0.500).

1.7 x 10-5 = x2/(0.500)
x = 2.9 x 10-3
Determine the pH of the solution. Since "x" represents the hydroxide ion concentration, we can convert it into pOH and than find the pH.
pOH = -log(2.9 x 10-3) = 2.54
pH = 14 - 2.54 = 11.46

Chem_Mod
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Re: pH of salt solutions

Postby Chem_Mod » Thu Jun 14, 2018 4:54 pm

You do not have to know how to do the calculations (14B), but to tell if the solution is basic or not, check if the ion that is formed from the salt is the conjugate base or the conjugate acid of the acid/base you know.


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