Hydrolysis Problems #11 - 20 | Calculations with salts of weak acids | |

Intro to Hydrolysis Calculations | Calculations with salts of weak bases | Acid Base Menu |

Note: the problems are all mixed up. Some are salts of weak acids, some are salts of weak bases.

**Problem #1:** CH_{3}NH_{2}, methyl amine is a weak base with a K_{b} of 4.38 x 10¯^{4}. What would be the pH of a solution of 0.350 M methyl ammonium chloride, CH_{3}NH_{3}^{+}Cl¯?

**Some discussion before the solution:**

1) This problem type seems to be seldom asked. Much more popular is giving you the K_{a} of an acid and then asking about pH of a solution formed from the salt of the acid. The problem above give you the K_{b} of a base and asks you for the pH of a salt of that base.

2) A bit of a warning: watch out, because there could be a temptation to teach the OTHER problem type, then test on this problem type. You have been warned!!

3) 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_{a} of CH_{3}NH_{3}^{+} (the acid) from the K_{b} of CH_{3}NH_{2} (the base).

4) See this tutorial for how the equation is derived. A reminder: the equation above applies to conjugate acid-base pairs. CH_{3}NH_{2} (the base) and CH_{3}NH_{3}^{+} (the acid) is the conjugate acid-base pair involved in the problem being discussed.

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

**Solution:**

1) Here is the relevant chemical equation for the above problem:

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

2) We must calculate the K_{a} for the above reaction. We will use the K_{b} of CH_{3}NH_{2} to do this:

1.00 x 10¯^{14}= (x) (4.38 x 10¯^{4})x = 1.00 x 10¯

^{14}/ 4.38 x 10¯^{4}= 2.2831 x 10¯^{11}

3) Please note that I left some guard digits on the K_{a} value. I will round off the final answer to the appropriate number of significant figures.

4) We now calculate the [H_{3}O^{+}] using the K_{a} expression:

2.2831 x 10¯^{11}= [(x) (x)] / (0.350 - x)neglect the minus x

x = 2.8268 x 10¯

^{6}M

6) Since we have the [H_{3}O^{+}], our last step will be to calculate the pH:

pH = - log 2.8268 x 10¯^{6}= 5.549

Note that, since this is the salt of a weak base (which is itself an acid), the calculated pH is an acidic value. This is as it should be!

**Problem #2:** What is the pH of 0.410 M methylammonium bromide, CH_{3}NH_{3}Br? (K_{b} of CH_{3}NH_{2} = 4.4 x 10¯^{4}.)

Note the differences in wording between this problem and the one just prior. In the prior problem, chloride was the spectator ion. In this problem, it is bromide. In this problem, the K_{b} value has been rounded off as well as the concentration being different.

**Solution:**

1) The relevant chemical equation is the hydrolysis of methyl amine:

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

2) We must calculate the K_{a} for the above reaction. We will use the K_{b} of CH_{3}NH_{2} to do this:

1.00 x 10¯^{14}= (x) (4.4 x 10¯^{4})x = 1.00 x 10¯

^{14}/ 4.4 x 10¯^{4}= 2.2727 x 10¯^{11}

3) We now calculate the [H_{3}O^{+}] using the K_{a} expression:

2.2727 x 10¯^{11}= [(x) (x)] / (0.410)x = 3.05257 x 10¯

^{6}M

6) Since we have the [H_{3}O^{+}], our last step will be to calculate the pH:

pH = - log 3.05257 x 10¯^{6}= 5.515Since the provided K

_{b}value had two sig figs, the best answer to report for the pH is 5.52.

**Problem #3:** What is the pH of a solution prepared by adding 0.750 g of ammonium chloride to 125 mL of water? (K_{b} of NH_{3} is 1.77 x 10¯^{5})

**Solution:**

1) Molarity first:

MV = mass / molar mass(x) (0.125 L) = 0.750 g / 53.4916 g/mol

x = 0.112167 M (we'll just keep a few guard digits and round off to a more appropriate value at the end)

2) The K_{a} for NH_{4}^{+}

K_{w}= K_{a}K_{b}1.00 x 10¯

^{14}= (K_{a}) (1.77 x 10¯^{5})K

_{a}= 5.65 x 10¯^{10}

3) Use the K_{a} expression:

(x) (x) 5.65 x 10¯ ^{10}=------------ 0.112167 x = 7.9608 x 10¯

^{6}M (this is the H^{+}conc)

4) Calculate the pH:

pH = -log 7.9608 x 10¯^{6}= 5.099

**Problem #4:** Calculate the pH of a 0.970 M aqueous solution of hydrazine hydrochloride (H_{2}NNH_{3}Cl). (For hydrazine, H_{2}NNH_{2}, K_{b} = 1.26 x 10¯^{6}.)

**Solution:**

1) Calculate the K_{a} of H_{2}NNH_{3}^{+}, the hydrazinium ion:

K_{a}K_{b}= K_{w}(K

_{a}) (1.26 x 10¯^{6}) = 1.00 x 10¯^{14}K

_{a}= 7.9365 x 10¯^{9}

2) Hydrazinium hydrolyzes as follows:

H_{2}NNH_{3}^{+}+ H_{2}O ⇌ H_{3}O^{+}+ H_{2}NNH_{2}

3) Solve the K_{a} expression for [H_{3}O^{+}]:

(x) (x) 7.9365 x 10¯ ^{9}=------------ 0.970 x = $\sqrt{\mathrm{[(7.9365\; x\; 10\xaf9)\; (0.970)]}}$

x = 0.00008774 M

4) The pH:

pH = -log 0.00008774 = 4.057

**Problem #5:** What is the pH of a 0.502 M solution of NH_{4}NO_{3}? The ionization constant of the weak base NH_{3} is 1.77 x 10¯^{5}

**Solution:**

1) The chemical equation of interest is this:

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

2) We need the K_{a} of the ammonium ion:

K_{w}= K_{a}K_{b}1.00 x 10¯

^{14}= (K_{a}) (1.77 x 10¯^{5})K

_{a}= 5.65 x 10¯^{10}

3) Calculate the hydrogen ion concentration, then pH:

5.65 x 10¯^{10}= [(x) (x)] / 0.502x = 1.684 x 10¯

^{5}MpH = - log 1.684 x 10¯

^{5}= 4.774

**Problem #6:** A 0.50 M solution of NaNO_{2} is prepared. Calculate the pH of this solution.

**Solution:**

1) The conjugate acid of NO_{2}¯ is HNO_{2}. Look up its K_{a}:

4.0 x 10¯^{4}

2) The nitrous anion behaves as a base in solution. Write the chemical reaction:

NO_{2}¯ + H_{2}O ⇌ HNO_{2}+ OH¯

3) Write the K_{b} expression for the nitrous anion and solve it for the hydroxide ion concentration:

(x) (x) 2.5 x 10¯ ^{11}=------------ 0.50 x = [OH¯] = 3.5355 x 10¯

^{6}M

4) pOH, then pH

pOH = -log 3.5355 x 10¯^{6}= 5.45pH = 14 - 5.45 = 8.55

**Problem #7:** Calculate the pH of a 0.360 M aqueous solution of ethylamine hydrochloride (C_{2}H_{5}NH_{3}Cl).
(For ethylamine, C_{2}H_{5}NH_{2}, K_{b} = 5.60 x 10¯^{4}.)

**Solution:**

1) The chemical reaction for the hydrolysis of ethylamine hydrochloride is:

C_{2}H_{5}NH_{3}^{+}+ H_{2}O ⇌ H_{3}O^{+}+ C_{2}H_{5}NH_{2}Note that the chloride ion (a spectator ion) has been eliminated.

2) A K_{a} expression for C_{2}H_{5}NH_{3}^{+} can be written:

[H _{3}O^{+}] [C_{2}H_{5}NH_{2}]K _{a}=--------------------------- [C _{2}H_{5}NH_{3}^{+}]

3) Let us solve the K_{a} expression:

(x) (x) 1.7857 x 10¯ ^{11}=------------ 0.360 x = 2.53545 x 10¯

^{6}M <--- this is the [H_{3}O^{+}]Note: the value for the K

_{a}came from this relationship:K_{a}K_{b}= K_{w}

4) Get the pH:

pH = -log 2.53545 x 10¯^{6}= 5.596

**Problem #8:** 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 the 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

**Problem #9:** Calculate the pH of a 0.200 M solution of ethanolammonium chloride, HOCH_{2}CH_{2}NH_{3}Cl. The weak base ethanolamine, HOCH_{2}CH_{2}NH_{2}, has a base ionization equilibrium constant (K_{b}) of 3.1 x 10¯^{5}.

**Solution:**

1) Ethanolammonium ion is the conjugate acid of the weak base ethanolamine, HOCH_{2}CH_{2}NH_{2}

2) Conjugate pairs have this relationship for their K's:

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

3) Calculate the K_{a} of the ethanolammonium ion:

(1 x 10¯^{14}) / (3.1 x 10¯^{5}) = 3.22581 x 10¯^{10}

4) Ethanolammonium ion hydrolyzes in water:

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

5) Solve the K_{a} expression for HOCH_{2}CH_{2}NH_{3}^{+}:

(x) (x) 3.22581 x 10¯ ^{10}=------------ 0.200 x = [H

_{3}O^{+}] = 8.0322 x 10¯^{6}M

6) Calculation of the pH:

pH = -log 8.0322 x 10¯^{6}pH = 5.10 (rounded to two sig figs)

**Problem #10:** Calculate the pH of a 0.100 M NaCN solution. The K_{a} for HCN, the conjugate acid of CN¯, equals 6.17 x 10¯^{10}

**Solution:**

1) Write the hydrolysis equation for the cyanide anion:

CN¯ + H_{2}O ⇌ OH¯ + HCN

2) Determine the K_{b} for CN¯:

K_{a}K_{b}= K_{w}(6.17 x 10¯

^{10}) (K_{b}) = 1.00 x 10¯^{14}K

_{b}= 1.62074 x 10¯^{5}

3) Solve the K_{b} expression for the hydroxide ion concentration:

(x) (x) 1.62074 x 10¯ ^{5}=------------ 0.100 x = 0.001273 M

4) First the pOH, then the pH:

pOH = -log 0.001273 = 2.895pH = 14 - 2.895 = 11.105

Hydrolysis Problems #11 - 20 | Calculations with salts of weak acids | |

Intro to Hydrolysis Calculations | Calculations with salts of weak bases | Acid Base Menu |