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**Problem #1:** 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?

**Some discussion before the solution:**

The solution to this type of problem depends on knowing that K_{w} = K_{a}K_{b}. We will use that equation to calculate the K_{b} of F¯ from the K_{a} 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 K_{b} 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 K_{b} value. I will round off the final answer to the appropriate number of significant figures.

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

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 - pOH = 14 - 5.212 = 8.788

**Problem #2:** 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:**

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}

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}= 5.36pH = 14 - pOH = 14 - 5.36 = 8.64

**Problem #3:** Calculate the pH of a 0.36 M CH_{3}COONa solution given the pK_{a} of acetic acid is 4.745.

**Solution:**

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

K_{a}= 10¯^{pKa}= 10¯^{4.745}= 1.80 x 10¯^{5}

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

1.00 x 10¯^{14}= (1.80 x 10¯^{5}) (x)x = 5.56 x 10¯

^{10}

Use the K_{b} 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}= 4.85pH = 14 - pOH = 14 - 4.85 = 9.15

Comment on the solution to Example #3:

You can use the pK_{a} to get the pK_{b}

4.745 + x = 14.000x = 9.255

Use the pK_{b} to get the K_{b}:

K_{b}= 10¯^{pKb}= 10¯^{9.255}= 5.56 x 10¯^{10}

Go to rest of solution, at "Use the K_{b} expression," just above.

**Problem #4:** 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:**

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

CN¯ + H_{2}O ⇔ HCN + OH¯

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.)

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

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

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

**Problem #5:** 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}) (x) = 1.00 x 10^{-14}x = 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

**Problem #6:** What is the pH of an aqueous solution of 0.620 M KNO_{2}?

Comment: to solve this problem, we need to know something more. What we need is the K_{a} for nitrous acid, HNO_{2}. 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 ionization equation for KNO_{2}:

NO_{2}¯ + H_{2}O <==> HNO_{2}+ OH¯

2) Use K_{a} of nitrous acid to calculate the K_{b} 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)] / 0.620x = 3.797 x 10

^{-6}M

4) Calculate pH

pOH = -log 3.797 x 10^{-6}= 5.42pH = 14.00 - 5.42 = 8.58

I originally answered this problem on Yahoo Answers and said that the K_{a} 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 K_{a} value, they were sure there was a way to solve the problem without recourse to the K_{a}.

OK then. Good luck with that.

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

**Problem #8:** Someday!

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