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To study the action of a buffer solution

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Introduction

EXPERIMENT ( 14 ) Topic : To study the action of a buffer solution Introduction : In this experiment, you are to compare the effects of adding small amounts of acid and alkali to buffered and unbuffered solutions of the same pH. You are provided with a buffer solution designed to maintain a pH of 7.0 at 25OC and some pure water which, if it is pure enough, should also have a pH of 7.0 at 25OC. To samples of these 2 liquids, you add small measured amounts of 0.1M NaOH and 0.1M HCl, measuring the pH at each addition. By comparing the pH changes in the 2 solutions, you can demonstrate the action of a buffer solution. Chemicals : 0.1M HCl(aq) , 0.1M NaOH(aq) , Distilled water , Buffer solution (pH 7.0) , Pure water (freshly-boiled distilled water) Procedures : Part A : Effect of NaOH on pH of buffer solution (pre-set at pH 7) 1. Fill a burette with 0.1M NaOH. 2. Using a measuring cylinder, put 25 cm3 of the buffer solution in a 50 cm3 beaker. 3. Rinse the pH meter electrode with distilled water from a wash-bottle, and put it into the beaker, making sure that the glass bulb is completely immersed. ...read more.

Middle

After adding NaOH into buufer solution, ethanoic acid will react with OH- to form salt and water. The equation is: CH3COOH + OH- CH3COO- + H2O The pH of the buffer solution only slightly rises from 7.0 to 7.4. The pH value can be calculated by this equation: pH = pKa + log [A-]/[HA] Since Ka is the equilibrium constant of the reaction, the pH value is only affected by the ratio of concentration of CH3COO- and CH3COOH. Because the ratio is in log form, the pH value will have a smaller change. 2. Explain, by using suitable equations, the effect of HCl on pH of buffer solution. After adding HCl into buufer solution, ethanoate will react with H+ to form ethanoic acid and water. The equation is: CH3COO- + H+CH3COOH + H2O The pH of the buffer solution only slightly drops from 7.2 to 6.4. The pH value can be calculated by this equation: pH = pKa + log [A-]/[HA] Since Ka is the equilibrium constant of the reaction, the pH value is only affected by the ratio of concentration of CH3COO- and CH3COOH. Because the ratio is in log form, the pH value will have a smaller change. ...read more.

Conclusion

Then, CH3COO- and H2O will be formed as a product. NaOH + CH3COOH CH3COO-Na+ + H2O Secondly, mix the sodium ethanoate solution and ethanoic acid together into a beaker. Put the data into this equation: pH = pKa + log [CH3COO-]/[ CH3COOH] 5.2 = -log (1.8x10-5) + log [CH3COO-]/1.0 5.2-4.74 = log [CH3COO-] 0.46 = log [CH3COO-] [CH3COO-] = 2.85M If 1 dm3 of the buffer solution is produced, the number of mole that CH3COO- needed =2.85(1) = 2.85 mole. Number of mole of CH3COO- = Number of mole of NaOH So the mass of NaOH that needed to use for making pH 5.2 buffer is 2.85 X (23+16+1) = 114g 10. Explain how could you measure Ka of 1.0M CH3COOH by using the following : (Given : 1.0M CH3COOH(aq) , 1.0M NaOH(aq) , pH meter) Firstly, add 1.0M NaOH(aq) into 1.0M CH3COOH. Then, CH3COO- and H2O will be formed as a product. NaOH + CH3COOH CH3COO-Na+ + H2O Secondly, mix the sodium ethanoate solution and ethanoic acid together into a beaker. The mixed solution is the buffer solution that we want to find its pH. Finally, put the electrode that belongs to pH meter into the buffer solution. The reading on the pH meter shows the pH of this buffer solution. ...read more.

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