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**Problem #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 = square root of [(5.65 x 10¯^{10}) (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

**Problem #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 = square root of [(2.70 x 10¯^{11}) (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

**Problem #3:**

Given the pK_{a} for ammonium ion is 9.26, 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¯^{9.26}K

_{a}= 5.4954 x 10¯^{10}

3) Write the dissociation equation and K_{a} expression:

NH_{4}^{+}⇌ H^{+}+ NH_{3}K

_{a}= ([H^{+}] [NH_{3}]) / [NH_{4}^{+}]

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

5.4954 x 10¯^{10}= [(x) (x)] / 0.100x = 7.4131 x 10¯

^{6}M

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

-log 7.4131 x 10¯^{6}= 5.13

There is a comment at the end of the calculation tutorial for salts of weak acid, just above the practice problems. Please read it if you skipped it before. Also, with specific reference to salts of weak bases

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Problems 1 & 2 are salt of weak acid problems. Please do them first.

3) 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.

4) 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.)

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