
For preparing a buffer of pH = 6 by mixing sodium acetate and acetic acid the ratio of the concentration of salt and acid should be ( $ {K_a} = {10^{ - 5}} $ ).
Answer
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Hint: In order to solve this question, we have to apply the formula of pH which will in terms of logarithm after putting the values given we will make the appropriate substitution and then take the antilog of it then we will get the ratio of the salt and the acid will be found.
Complete step by step solution:
To solve this question first, we will have to know what PH is.
PH is a quantitative measure of the acidity or basicity of aqueous or other liquid solutions. The term, widely used in chemistry, translates the values of the concentration of the hydrogen ion which ordinarily ranges between about 1 and 10−14 gram equivalents per litre into numbers between 0 and 14.
In pure water that is neutral which is neither acidic nor basic. The concentration of the hydrogen ion is 10−7 gram equivalents per litre, which equals to a pH of 7. A solution with a pH less than 7 regarded as acidic. A solution with a pH greater than 7 is considered alkaline.
Formula for calculating the pH;
$ pH = - \log {K_a} + \log \dfrac{{[salt]}}{{[acid]}} $
Since the value of pH and $ {K_a} $ are given in question so putting these values in the given formula:
$ 6 = - \log {10^{ - 5}} + \log \dfrac{{[salt]}}{{[acid]}} $
From the property of logarithm $ [a\log b = \log {b^a}] $ :
$ 6 = 5 + \log \dfrac{{[salt]}}{{[acid]}} $
On further solving:
$ 1 = \log \dfrac{{[salt]}}{{[acid]}} $
Taking antilog on both side:
$ \dfrac{{10}}{1} = \dfrac{{[salt]}}{{[acid]}} $
So this will be our final solution. The ratio of salt to the acid will be 10:1.
Note:
The pH is usually measured with a pH meter, which converts into pH readings the difference in electromotive force which is also called electrical potential or voltage between electrodes placed in the solution to be tested. Fundamentally, a pH meter consists of a voltmeter attached to a pH responsive electrode and a reference constant electrode.
Complete step by step solution:
To solve this question first, we will have to know what PH is.
PH is a quantitative measure of the acidity or basicity of aqueous or other liquid solutions. The term, widely used in chemistry, translates the values of the concentration of the hydrogen ion which ordinarily ranges between about 1 and 10−14 gram equivalents per litre into numbers between 0 and 14.
In pure water that is neutral which is neither acidic nor basic. The concentration of the hydrogen ion is 10−7 gram equivalents per litre, which equals to a pH of 7. A solution with a pH less than 7 regarded as acidic. A solution with a pH greater than 7 is considered alkaline.
Formula for calculating the pH;
$ pH = - \log {K_a} + \log \dfrac{{[salt]}}{{[acid]}} $
Since the value of pH and $ {K_a} $ are given in question so putting these values in the given formula:
$ 6 = - \log {10^{ - 5}} + \log \dfrac{{[salt]}}{{[acid]}} $
From the property of logarithm $ [a\log b = \log {b^a}] $ :
$ 6 = 5 + \log \dfrac{{[salt]}}{{[acid]}} $
On further solving:
$ 1 = \log \dfrac{{[salt]}}{{[acid]}} $
Taking antilog on both side:
$ \dfrac{{10}}{1} = \dfrac{{[salt]}}{{[acid]}} $
So this will be our final solution. The ratio of salt to the acid will be 10:1.
Note:
The pH is usually measured with a pH meter, which converts into pH readings the difference in electromotive force which is also called electrical potential or voltage between electrodes placed in the solution to be tested. Fundamentally, a pH meter consists of a voltmeter attached to a pH responsive electrode and a reference constant electrode.
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