### Calculate the pH of a solution of the salt, given the Ka (or the pKa) of the acid

Problem #1: HF, hydrofluoric acid is a weak acid with a Ka of 3.55 x 10¯4. What would be the pH of a solution of 1.34 M sodium fluoride?

Some discussion before the solution:

The solution to this type of problem depends on knowing that Kw = KaKb. We will use that equation to calculate the Kb of F¯ from the Ka of HF.

See this tutorial for how the equation is derived. A reminder: the equation above applies to conjugate acid-base pairs. HF and F¯ is the conjugate acid-base pair involved in the problem being discussed.

Another important point to know is how salts hydrolyze. In the problem being examined, we have the salt of a weak acid hydrolyzing. See this tutorial for more explanation.

Solution.

We will calculate the Kb of F¯:

1.00 x 10¯14 = (3.55 x 10¯4) (x)

x = 1.00 x 10¯14 / 3.55 x 10¯4 = 2.8169 x 10¯11

Please note that I left some guard digits on the Kb value. I will round off the final answer to the appropriate number of significant figures.

We now need to calculate the [OH¯] using the Kb expression:

2.8169 x 10¯11 = [(x) (x)] / (1.34 - x)

neglect the minus x

x = 6.1438 x 10¯6 M

Since this is the [OH¯], we will calculate the pOH and thence to the pH:

pOH = - log 6.1438 x 10¯6 = 5.212

pH = 14 - pOH = 14 - 5.212 = 8.788

Problem #2: The Ka of formic acid (HCOOH) is 1.8 x 10¯4. What is the pH of 0.35 M solution of sodium formate (NaHCOO)?

Solution:

Calculate the Kb of the formate (HCOO¯) ion:

1.00 x 10¯14 = (1.8 x 10¯4) (x)

x = 1.00 x 10¯14 / 1.8 x 10¯4 = 5.56 x 10¯11

Calculate the [OH¯]:

5.56 x 10¯11 = [(x) (x)] / (0.35 - x)

neglect the minus x

x = 4.41 x 10¯6 M

Calculate pOH, then pH:

pOH = - log 4.41 x 10¯6 = 5.36

pH = 14 - pOH = 14 - 5.36 = 8.64

Problem #3: Calculate the pH of a 0.36 M CH3COONa solution given the pKa of acetic acid is 4.745.

Solution:

Using the pKa, calculate the Ka of acetic acid:

Ka = 10¯pKa = 10¯4.745 = 1.80 x 10¯5

Using the Ka of the acid, find the Kb of the base (the CH3COONa):

1.00 x 10¯14 = (1.80 x 10¯5) (x)

x = 5.56 x 10¯10

Use the Kb expression to calculate the [OH¯]:

5.56 x 10¯10 = [(x) (x)] / (0.36 - x)

neglect the minus x

x = 1.415 x 10¯5 M (I threw in a guard digit.)

Calculate pOH, then pH:

pOH = - log 1.415 x 10¯5 = 4.85

pH = 14 - pOH = 14 - 4.85 = 9.15

Comment on the solution to Example #3:

You can use the pKa to get the pKb

4.745 + x = 14.000

x = 9.255

Use the pKb to get the Kb:

Kb = 10¯pKb = 10¯9.255 = 5.56 x 10¯10

Go to rest of solution, at "Use the Kb expression," just above.

Problem #4: What is the pH of a 0.0510 molar solution of salt NaCN (the Ka for HCN is 6.166 x 10¯10)?

Solution:

CN¯ is the salt of a weak acid and hydrolyzes thusly:

CN¯ + H2O ⇌ HCN + OH¯

Determine the Kb for this reaction (using Kw = KaKb):

1.00 x 10¯14 = (6.166 x 10¯10) (x)

x = 1.6218 x 10¯5 (I kept a couple guard digits.)

Write the Kb expression and solve for [OH-]:

Kb = ([HCN] [OH¯)] / [CN¯]

1.6218 x 10¯5 = [(x) (x)] / 0.0510

x = 9.0946 x 10¯4 M

Determine pOH (with negative log) and then pH (using pH + pOH = 14):

-log 9.0946 x 10¯4 = 3.0412

pH = 14 - 3.0412 = 10.959 (to three sig figs)

Problem #5: A certain weak acid, HA, with a Ka value of 5.61 x 10-6, is titrated with NaOH. A solution is made by mixing 8.00 mmol of HA and 3.00 mmol of the strong base. The resulting pH is 5.03. More strong base is added until the equivalence point is reached. What is the pH of this solution at the equivalence point if the total volume is 41.0 mL?

Solution:

Comment: The information about the pH being 5.03 is not needed. It's there as a distractor. Also, I kept some guard digits.

1) Caculate concentration of salt:

The 8.00 mmole of HA reacted in a 1:1 molar ratio with the NaOH to make 8.00 mmol of NaA.

8.00 mmol / 41.0 mL = 0.195 M

Reminder: mmol means millimole.

2) Calculate Kb of A¯:

(5.61 x 10-6) (x) = 1.00 x 10-14

x = 1.7825 x 10-9

3) Calculate [OH¯]:

A¯ + H2O ⇌ HA + OH¯

1.7825 x 10-9 = [(x) (x)] / 0.195

x = 1.864 x 10-5 M

4) Calculate pH

pOH = -log 1.864 x 10-5 = 4.730

pH = 14.000 - 4.730 = 9.270

Problem #6: What is the pH of an aqueous solution of 0.620 M KNO2?

Comment: to solve this problem, we need to know something more. What we need is the Ka for nitrous acid, HNO2. When looking around the Internet, there are several values used in various problems. This problem on Yahoo Answers provides the value I will use.

1) Write the ionization equation for KNO2:

NO2¯ + H2O ⇌ HNO2 + OH¯

2) Use Ka of nitrous acid to calculate the Kb for nitrous ion:

(4.3 x 10-4) (x) = 1.0 x 10-14

x = 2.3256 x 10-11 (I'll round off at the end.)

3) Calculate [OH¯]:

2.3256 x 10-11 = [(x) (x)] / 0.620

x = 3.797 x 10-6 M

4) Calculate pH

pOH = -log 3.797 x 10-6 = 5.42

pH = 14.00 - 5.42 = 8.58

I originally answered this problem on Yahoo Answers and said that the Ka for the acid was necessary and proceeded with the above explanation. The person gave me best answer, but commented that, since the question lacked the Ka value, they were sure there was a way to solve the problem without recourse to the Ka.

OK then. Good luck with that.

Problem #7: Problem 2 (under Practice Problems) in this tutorial is another problem of this type.

Problem #8: What is the pH for a 0.065 M barium cyanate, Ba(CNO)2 solution?

Solution:

We need a Ka for cyanic acid. Its value is 3.5 x 10¯4 (looked up on ye olde Internet).

1) The key reaction is this:

CNO¯(aq) + H2O(l) ⇌ HCNO + OH¯

2) Here's the relevant Kb of CNO¯ formulation:

Kb = [HCNO] [OH¯] / [CNO¯]

The [CNO¯] = 0.065 x 2 = 0.13 M

3) Since this is a Kb problem, we need its value. We use the Ka of cyanic acid for that:

Kw = Ka * Kb

1.0 x 10¯4 = (3.5 x 10¯4) (Kb)

Kb = 2.857 x 10¯11 <--- I won't round off until the end

4) Now, we calculate the hydroxide conc:

2.857 x 10¯11 = [(x) (x)] / 0.13

x = 1.926136 x 10¯6 M

5) I will next calculate the pOH and get the pH from that value.

pOH = -log 1.926136 x 10¯6 = 5.7153

pH = 14 - 5.7153 = 8.2847

8.28 is an appropriate value to report.