Hydrolysis Problems #1 - 10 | Calculations with salts of weak acids | ||

Hydrolysis Problems #11 - 20 | Intro to Hydrolysis Calculations | Acid Base Menu |

Note: in the first four problems, I give the K_{a} of the conjugate acid (for example, the ammonium ion in Example #1). Often, these problems are given with the K_{b} of the base and you have to calculate the value of the K_{a}. You do so with this equation:

K_{a}K_{b}= K_{w}

You will see such a situation starting in the fifth example as well as scattered through the additional problems.

**Example #1:** What is the pH of a 0.0500 M solution of ammonium chloride, NH_{4}Cl. K_{a} = 5.65 x 10¯^{10}.

**Solution:**

1) Here is the chemical reaction (net ionic) for the hydrolysis of NH_{4}Cl:

NH_{4}^{+}+ H_{2}O ⇌ NH_{3}+ H_{3}O^{+}

2) Here is the K_{a} expression for NH_{4}^{+}:

[NH _{3}] [H_{3}O^{+}]K _{a}=------------------ [NH _{4}^{+}]

3) We can then substitute values into the K_{a} expression in the normal manner:

(x) (x) 5.65 x 10¯ ^{10}=------------ 0.0500 - x

4) Ignoring the minus x in the usual manner, we proceed to sove for the hydronium ion concentration:

x = $\sqrt{\mathrm{(5.65\; x\; 10\xaf10)\; (0.0500)}}$x = 5.32 x 10¯

^{6}M = [H_{3}O^{+}]

5) We then calculate the pH directly from the [H_{3}O^{+}] value:

pH = - log 5.32 x 10¯^{6}= 5.274

**Example #2:** What is the pH of a 0.100 M solution of methyl ammonium chloride (CH_{3}NH_{3}Cl). K_{a} of the methyl ammonium ion (CH_{3}NH_{3}^{+} = 2.70 x 10¯^{11})

**Solution:**

1) Here is the chemical reaction (net ionic) for the hydrolysis of CH_{3}NH_{3}^{+}:

CH_{3}NH_{3}^{+}+ H_{2}O ⇌ CH_{3}NH_{2}+ H_{3}O^{+}

2) Here is the K_{a} expression for CH_{3}NH_{3}^{+}:

[CH _{3}NH_{2}] [H_{3}O^{+}]K _{a}=----------------------- [CH _{3}NH_{3}^{+}]

3) We can then substitute values into the K_{a} expression in the normal manner:

(x) (x) 2.70 x 10¯ ^{11}=----------- 0.100 - x

4) Ignoring the minus x in the usual manner, we proceed to sove for the hydronium ion concentration:

x = $\sqrt{\mathrm{(2.70\; x\; 10\xaf11)\; (0.100)}}$x = 1.64 x 10¯

^{6}M = [H_{3}O^{+}]

5) We then calculate the pH directly from the [H_{3}O^{+}] value:

pH = - log 1.64 x 10¯^{6}= 5.784

**Example #3:** Given the pK_{a} for ammonium ion is 9.248, what is the pH of 1.00 L of solution which contains 5.45 g of NH_{4}Cl (the molar mass of NH_{4}Cl = 54.5 g mol¯^{1}.)

**Solution:**

1) Determine molarity of the solution:

5.45 g / 54.5 g mol¯^{1}= 0.100 mol0.100 mol / 1.00 L = 0.100 M

2) Determine K_{a} for NH_{4}Cl:

K_{a}= 10¯^{pKa}= 10¯^{9.248}K

_{a}= 5.64937 x 10¯^{10}

3) Write the equation for the hydrolysis of the ammonium ion and K_{a} expression:

NH_{4}^{+}+ H_{2}O ⇌ H_{3}O^{+}+ NH_{3}

[H _{3}O^{+}] [NH_{3}]K _{a}=----------------- [NH _{4}^{+}]

4) Insert values into K_{a} expression and solve:

(x) (x) 5.64937 x 10¯ ^{10}=-------- 0.100 x = 7.51623 x 10¯

^{6}M

5) Take negative log of this value (this value being the [H_{3}O^{+}]) for the pH:

pH = -log 7.51623 x 10¯^{6}= 5.124

**Example #4:** Aniline is a weak organic base with the formula C_{6}H_{5}NH_{2}. The anilinium ion has the formula C_{6}H_{5}NH_{3}^{+} and its K_{a} = 2.3 x 10¯^{5}.

a) write the chemical reaction showing the hydrolysis of anilinium ion.

b) calculate the pH of a 0.0400 M solution of anilinium ion.

**Solution:**

1) The chemical reaction for the hydrolysis is:

C_{6}H_{5}NH_{3}^{+}+ H_{2}O ⇌ C_{6}H_{5}NH_{2}+ H_{3}O^{+}

2) The calculations for the pH are:

(x) (x) 2.3 x 10¯ ^{5}=---------------- 0.0400 - x x = $\sqrt{\mathrm{(2.3\; x\; 10\xaf5)\; (0.0400)}}$

x = 9.6 x 10¯

^{4}M = [H_{3}O^{+}]pH = - log 9.6 x 10¯

^{6}= 3.02

**Example #5:** Pyridine (C_{5}H_{5}N) is a weak base and reacts with HCl as follows:

C_{5}H_{5}N + HCl ---> C_{5}H_{5}NH^{+}+ Cl¯

What is the pH of a 0.015 M solution of the pyridinium ion (C_{5}H_{5}NH^{+})? The K_{b} for pyridine is 1.6 x 10¯^{9}. (Hint: calculate the K_{a} for the pyridinium ion and use it in the calculation.)

**Solution:**

1) First, we calculate the K_{a}:

K

_{a}K_{b}= K_{w}

1.00 x 10¯ ^{14}K _{a}=-------------- 1.6 x 10¯ ^{9}K

_{a}= 6.25 x 10¯^{6}

2) Now, the solution follows the pattern outlined in the tutorial:

C_{5}H_{5}NH^{+}+ H_{2}O ⇌ H_{3}O^{+}+ C_{5}H_{5}N

(x) (x) 6.25 x 10¯ ^{6}=-------- 0.015 x = $\sqrt{\mathrm{(6.25\; x\; 10\xaf6)\; (0.015)}}$

x = 3.06 x 10¯

^{4}M <--- this is the [H_{3}O^{+}]pH = - log 3.06 x 10¯

^{6}= 3.51

**Bonus Example:** The pH of a 0.160 M CH_{3}NH_{3}Cl solution is 5.500. What is the value of K_{b} for CH_{3}NH_{2}?

Solution technique: we will determine the K_{a} of CH_{3}NH_{3}Cl (since that is what we have data for). Based on the fact that CH_{3}NH_{2} is the conjugate base, we will use K_{a}K_{b} = K_{w} to get the K_{b}.

**Solution:**

1) Here's the K_{a} expression we are interested in:

[H _{3}O^{+}] [CH_{3}NH_{2}]K _{a}=------------------------ [CH _{3}NH_{3}^{+}]

2) It comes from this reaction, the hydrolysis of CH_{3}NH_{3}^{+}:

CH_{3}NH_{3}^{+}+ H_{2}O ⇌ H_{3}O^{+}+ CH_{3}NH_{2}

3) The concentrations of two components of the K_{a} expression come from the pH:

H_{3}O^{+}= 10¯^{5.500}= 3.16228 x 10¯^{6}MThis is also the concentration of the CH

_{3}NH_{2}.

4) Solve for the K_{a}:

(3.16228 x 10¯ ^{6}) (3.16228 x 10¯^{6})K _{a}=----------------------------------------- 0.160 K

_{a}= 6.25 x 10¯^{11}

5) Solve for the K_{b}:

K_{a}K_{b}= K_{w}(6.25 x 10¯

^{11}) (K_{b}) = 1.00 x 10¯^{14}K

_{b}= 1.6 x 10¯^{4}

Hydrolysis Problems #1 - 10 | Calculations with salts of weak acids | ||

Hydrolysis Problems #11 - 20 | Intro to Hydrolysis Calculations | Acid Base Menu |