Question

Consider how to prepare a buffer solution with pH =
**3.66** (using one of the weak acid/conjugate base
systems shown here) by combining **1.00** L of a
**0.305**-M solution of weak acid with
**0.271** M **sodium
hydroxide**.

Weak Acid | Conjugate Base | K_{a} |
pK_{a} |
---|---|---|---|

HNO |
NO |
4.5 x 10 |
3.35 |

HClO |
ClO |
3.5 x 10 |
7.46 |

HCN |
CN |
4.0 x 10 |
9.40 |

How many L of the **sodium hydroxide** solution would
have to be added to the acid solution of your choice?

L

Answer #1

**we should chose the acid whose pKa is closer to required
pH**

**So I chose HNO2/NO2- buffer**

**Let volume of NaOH added be V L**

**so, mol of NaOH added = 0.271 M * V L = 0.271*V
mol**

**mol of HNO2 present initially = 0.305 M * 1 L = 0.305
mol**

**both will react to form NO2-**

**mol of NO2- formed = 0.271*V mol**

**mol of HNO2 remaining = 0.305 - 0.271*V**

**use:**

**pH = pKa + log {[NO2-]/[HNO2]}**

**3.66 = 3.35 + log (0.271*V / (0.305 -
0.271*V))**

**log (0.271*V / (0.305 - 0.271*V)) = 0.31**

**0.271*V / (0.305 - 0.271*V) = 2.042**

**0.271*V = 0.623 - 0.553*V**

**V = 0.756 L**

**Answer: 0.756 L**

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Weak Acid
Conjugate Base
Ka
pKa
HNO2
NO2-
4.5 x 10-4
3.35
HClO
ClO-
3.5 x 10-8
7.46
HCN
CN-
4.0 x 10-10
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pKa
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