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 K_{a} for HCN is 6.166 x 10¯^{10})?

**Solution:**

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

CN¯ + H_{2}O ⇌ HCN + OH¯

2) Determine the K_{b} for this reaction (using K_{w} = K_{a}K_{b}):

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 K_{b} expression and solve for [OH¯]:

K_{b}= ([HCN] [OH¯)] / [CN¯]1.6218 x 10¯

^{5}= [(x) (x)] / 0.0510x = 9.0946 x 10¯

^{4}M

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

pH = −log 9.0946 x 10¯^{4}= 3.0412pH = 14 − 3.0412 = 10.959 (to three sig figs)

**Problem #12:** The K_{a} 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 K_{b} 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}= 5.36pH = 14 − 5.36 = 8.64

**Problem #13:** Calculate the pH of a 0.36 M CH_{3}COONa solution given the pK_{a} of acetic acid is 4.752.

**Solution:**

1) Using the pK_{a}, calculate the K_{a} of acetic acid:

K_{a}= 10¯^{pKa}= 10¯^{4.752}= 1.77 x 10¯^{5}

2) Using the K_{a} of the acid, find the K_{b} of the base (the CH_{3}COONa):

1.00 x 10¯^{14}= (1.77 x 10¯^{5}) (K_{b})K

_{b}= 5.65 x 10¯^{10}

3) Use the K_{b} 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 kept a guard digit.)

4) Calculate pOH, then pH:

pOH = −log 1.426 x 10¯^{5}= 4.84pH = 14 − 4.84 = 9.16

5) Comment on the start of the above solution:

You can use the pK_{a}to first get the pK_{b}:4.752 + pK_{b}= 14.000pK

_{b}= 9.248Use the pK

_{b}to get the K_{b}:K_{b}= 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 (pK_{b} = 5.80)?

**Solution:**

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

_{b}of codeine equals 5.80. The pK_{a}of codeine HCl = 8.20The chemical reaction is this:

CodeineH^{+}Cl¯ + H_{2}O ⇌ Codeine + H_{3}O^{+}+ Cl¯The Cl¯ is a spectator ion.

We need the K

_{a}, which is 6.3 x 10¯^{9}(I skipped a couple steps: convert pK

_{b}to K_{b}, use K_{a}K_{b}= K_{w})6.3 x 10¯

^{9}= [(x) (x)] / 0.064x = 2.0 x 10¯

^{5}M (this is the H_{3}O^{+}concentration)pH = 4.70

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

**Solution:**

1) The key reaction is this:

CNO¯(aq) + H_{2}O(ℓ) ⇌ HCNO + OH¯

2) Here's the relevant K_{b} of CNO¯ formulation:

K_{b}= [HCNO] [OH¯] / [CNO¯]

3) Since this is a K_{b} problem, we need its value. We use the K_{a} of cyanic acid for that (3.5 x 10¯^{4} was looked up on ye olde Internet).

K_{w}= K_{a}K_{b}1.0 x 10¯

^{4}= (3.5 x 10¯^{4}) (K_{b})K

_{b}= 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)] / 0.13x = 1.926136 x 10¯

^{6}MNote: the [CNO¯] of 0.13 M comes from the fact that every Ba(CNO)

_{2}releases two CNO¯ to the solution. 0.065 times 2 equals 0.13.

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

pOH = −log 1.926136 x 10¯^{6}= 5.7153pH = 14 − 5.7153 = 8.2847

8.28 is an appropriate value to report.

**Problem #16:** HF, hydrofluoric acid is a weak acid with a K_{a} 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 K_{b} of F¯:

1.00 x 10¯^{14}= (3.55 x 10¯^{4}) (K_{b})K

_{b}= 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 K_{b} 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 K_{b} 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}= 5.212pH = 14 − 5.212 = 8.788

**Problem #17:** A certain weak acid, HA, with a K_{a} 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 K_{b} of A¯:

(5.61 x 10^{-6}) (K_{b}) = 1.00 x 10^{-14}K

_{b}= 1.7825 x 10^{-9}

3) Calculate [OH¯]:

A¯ + H_{2}O ⇌ HA + OH¯1.7825 x 10

^{-9}= [(x) (x)] / 0.195x = 1.864 x 10

^{-5}M

4) Calculate pH

pOH = −log 1.864 x 10^{-5}= 4.730pH = 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 |