Hydrolysis calculations: Problems #11 - 20

Hydrolysis Problems #1 - 10     Calculations with salts of weak acids
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Note: the problems are all mixed up. Some are salts of weak acids, some are salts of weak bases.

Problem #11: What is the pH of a 0.0510 molar solution of salt NaCN (the Ka for HCN is 6.166 x 10¯10)?

Solution:

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

CN¯ + H2O ⇌ HCN + OH¯

2) 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.)

3) 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

4) 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 #12: 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:

1) 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

2) Calculate the [OH¯]:

5.56 x 10¯11 = [(x) (x)] / (0.35 - x)

neglect the minus x

x = 4.41 x 10¯6 M

3) Calculate pOH, then pH:

pOH = - log 4.41 x 10¯6 = 5.36

pH = 14 - 5.36 = 8.64


Problem #13: Calculate the pH of a 0.36 M CH3COONa solution given the pKa of acetic acid is 4.752.

Solution:

1) Using the pKa, calculate the Ka of acetic acid:

Ka = 10¯pKa = 10¯4.752 = 1.77 x 10¯5

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

1.00 x 10¯14 = (1.77 x 10¯5) (Kb)

Kb = 5.65 x 10¯10

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

5.65 x 10¯10 = [(x) (x)] / (0.36 - x)

neglect the minus x

x = 1.426 x 10¯5 M (I threw in a guard digit.)

4) Calculate pOH, then pH:

pOH = - log 1.426 x 10¯5 = 4.84

pH = 14 - 4.84 = 9.16

5) Comment on the start of the above solution:

You can use the pKa to first get the pKb:
4.752 + pKb = 14.000

pKb = 9.248

Use the pKb to get the Kb:

Kb = 10¯pKb = 10¯9.248 = 5.65 x 10¯10

Continue on at step 3, just above.


Problem #14: Codeine is a narcotic pain reliever that forms a salt with HCl. What is the pH of 0.064 codeine hydrochloride (pKb = 5.80)?

Solution:

Codeine is a weak base. Codeine HCl is the salt of a weak base and so would create an acidic solution.

pKb of codeine equals 5.80. The pKa of codeine HCl = 8.20

The chemical reaction is this:

CodeineH+ + H2O ⇌ Codeine + H3O+

The Cl¯ is a spectator ion.

We need the Ka, which is 6.3 x 10¯9

6.3 x 10¯9 = [(x) (x)] / 0.064

x = 2.0 x 10¯5 M (this is the H3O+ concentration)

pH = 4.70


Problem #15: What is the pH for a 0.065 M barium cyanate, Ba(CNO)2 solution?

Solution:

We need a Ka for cyanic acid (HCNO). 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 = KaKb

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.


Problem #16: 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?

Solution.

1) We will calculate the Kb of F¯:

1.00 x 10¯14 = (3.55 x 10¯4) (Kb)

Kb = 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.

2) 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

3) 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 #17: 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) (Kb) = 1.00 x 10-14

Kb = 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


Hydrolysis Problems #1 - 10     Calculations with salts of weak acids
Intro to Hydrolysis Calculations     Calculations with salts of weak bases     Acid Base Menu