sarah=] 2012-4-23 07:55 PM

chungkin81 2012-4-23 08:26 PM

You can consider this question in buffer:

Let the volume of HA added be V, the volume of NaOH added should also be V.

Concentration of HA after mixing = (0.2 x V - 0.1 x V) / 2V = 0.05 M

Concentration of NaA formed after mixing = 0.1 x V / 2V = 0.05 M

By pH = pKa + log [salt] / [ acid]

= -log (1E-6) + log (0.05/0.05)

= 6

Let the volume of HA added be V, the volume of NaOH added should also be V.

Concentration of HA after mixing = (0.2 x V - 0.1 x V) / 2V = 0.05 M

Concentration of NaA formed after mixing = 0.1 x V / 2V = 0.05 M

By pH = pKa + log [salt] / [ acid]

= -log (1E-6) + log (0.05/0.05)

= 6

sarah=] 2012-4-23 09:33 PM

thank you. if you don't mind, plaese also help me with this question:

A solution of pH 3.8 is to be made by mixing 1 M solution of ethanoic acid (pKa=4.74) and sodium ethanoate. what proportion by volume should the solutions be mixed?

if possible, please explain it to me without using -Henderson hassalbalch equation.

Thanks

A solution of pH 3.8 is to be made by mixing 1 M solution of ethanoic acid (pKa=4.74) and sodium ethanoate. what proportion by volume should the solutions be mixed?

if possible, please explain it to me without using -Henderson hassalbalch equation.

Thanks

chungkin81 2012-4-24 04:54 PM

In fact, this question is simpler to consider by applying the concept of buffer.

As ethanoic acid is a weak acid, it dissociates in water slightly and establish an equilibrium as follows:

CH3COOH (aq) + H2O (l) <==> CH3COO- (aq) + H3O+ (aq)

Sodium ethanoate is a salt, it will dissociate in water as follows:

CH3COONa (aq) --> CH3COO- (aq) + Na+ (aq)

By the Le Chatelier's principle, the ethanoate ions from sodium ethanoate will shift the equilibrium position of the acid dissociation to the left.

Hence, the concentration of CH3COO- is largely comes from sodium ethanoate and

the concentration of CH3COOH is largely comes from ethanoic acid only

Let the volume of 1M ethanoic acid added to be V1 and the volume of 1M sodium ethanoate to be V2.

After mixing the two solutions, the total volume of the mixture is V1 + V2.

Applying the equation,

pH = pKa + log [CH3COO-] / [ CH3COOH]

3.8 = 4.76 + log [CH3COO-] / [CH3COOH]

-0.94 = log [CH3COO-] / [CH3COOH]

[CH3COO-] / [CH3COOH] = 10^(-0.94) = 0.1148

[1 x V1 / (V1 + V2)] / [ 1 X V2 / (V1 + V2)] = 0.1148

V1 / V2 = 0.1148

V1 = 0.1148 V2

So, the volume ratio of ethanoic acid to sodium ethanoate is 0.1148 : 1

As ethanoic acid is a weak acid, it dissociates in water slightly and establish an equilibrium as follows:

CH3COOH (aq) + H2O (l) <==> CH3COO- (aq) + H3O+ (aq)

Sodium ethanoate is a salt, it will dissociate in water as follows:

CH3COONa (aq) --> CH3COO- (aq) + Na+ (aq)

By the Le Chatelier's principle, the ethanoate ions from sodium ethanoate will shift the equilibrium position of the acid dissociation to the left.

Hence, the concentration of CH3COO- is largely comes from sodium ethanoate and

the concentration of CH3COOH is largely comes from ethanoic acid only

Let the volume of 1M ethanoic acid added to be V1 and the volume of 1M sodium ethanoate to be V2.

After mixing the two solutions, the total volume of the mixture is V1 + V2.

Applying the equation,

pH = pKa + log [CH3COO-] / [ CH3COOH]

3.8 = 4.76 + log [CH3COO-] / [CH3COOH]

-0.94 = log [CH3COO-] / [CH3COOH]

[CH3COO-] / [CH3COOH] = 10^(-0.94) = 0.1148

[1 x V1 / (V1 + V2)] / [ 1 X V2 / (V1 + V2)] = 0.1148

V1 / V2 = 0.1148

V1 = 0.1148 V2

So, the volume ratio of ethanoic acid to sodium ethanoate is 0.1148 : 1

病切斯特魚拉特 2012-5-6 07:40 PM

*** 作者被禁止或刪除 內容自動屏蔽 ***

頁:
**[1]**

查看完整版本: **Aqueous equilibria**

Copyright © 2003-2019 香港討論區