What ratio of acetic acid to sodium acetate concentration is needed to achieve a buffer whose $pH$ is $5.70$? The dissociation constant of acetic acid is $1.8 \times {10^{ - 5}}$.
Answer
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Hint: The dissociation constant can be used for determining the ratio of acid to salt concentration. Using $pH$ of the given solution this can be determined from the given equation formulated for a mixture of acid and salt. This ratio can easily be determined by solving the equation.
Complete step by step answer:
The $pH$ of the mixture of the solution formed with an acid and a salt needs to be $5.70$. It is given that the dissociation constant of acetic acid is $1.8 \times {10^{ - 5}}$. The dissociation constant is important in this case as there is presence of both acid and salt in the given solution. In this case,
${K_a} = 1.8 \times {10^{ - 5}}$
The process of determining the $pH$ value of the solution is possible using the equation:
$pH = p{K_a} + \log \dfrac{{\left[ {Salt} \right]}}{{\left[ {Acid} \right]}}$
Here the $\dfrac{{\left[ {Salt} \right]}}{{\left[ {Acid} \right]}}$ is the ratio of the sodium salt with respect to the acetic acid.
In this case from ${K_a} = 1.8 \times {10^{ - 5}}$ the value of $p{K_a}$ can be determined. The value of $p{K_a}$ is the negative logarithm of dissociation constant value. The value of $p{K_a}$ can be calculated as:
\[p{K_a} = - \log \left( {1.8 \times {{10}^{ - 5}}} \right)\]
$ \Rightarrow p{K_a} = - \{ \log (1.8 \times {10^{ - 5}})\} $
$ \Rightarrow p{K_a} = - \{ \log (1.8) + \log ({10^{ - 5}})\} $ The logarithms are separated
$ \Rightarrow p{K_a} = - \{ \log \left( {1.8} \right) - 5\log (10)\} $ Power is brought as the coefficient
$ \Rightarrow p{K_a} = - \{ 0.255 - 5\} $ Putting the values of logarithm and log10=1 $ \Rightarrow p{K_a} = - \{ - 4.745\} $
Therefore, the value $p{K_a} = 4.745$
Hence from the given equation, putting the values of $pH$ given and the $p{K_a}$ calculated we get,
$5.7 = 4.745 + \log \dfrac{{\left[ {Salt} \right]}}{{\left[ {Acid} \right]}}$
This shows that, \[\log \dfrac{{\left[ {Salt} \right]}}{{\left[ {Acid} \right]}} = 5.7 - 4.745\]
Hence, we get the $\log \dfrac{{\left[ {Salt} \right]}}{{\left[ {Acid} \right]}} = 0.955$
Removing log on both sides we get, $\dfrac{{\left[ {Salt} \right]}}{{\left[ {Acid} \right]}} = 9.02$
In the given condition the ratio of the mixture required is $9.02$.
Note: The specific concentration of acid and the salt can be mixed in a solution to attain a particular $pH$ level. Thus, by adding a specific level of acid and salt a predetermined $pH$ can be attained. This is the importance of the specific components resulting in the particular $pH$.
Complete step by step answer:
The $pH$ of the mixture of the solution formed with an acid and a salt needs to be $5.70$. It is given that the dissociation constant of acetic acid is $1.8 \times {10^{ - 5}}$. The dissociation constant is important in this case as there is presence of both acid and salt in the given solution. In this case,
${K_a} = 1.8 \times {10^{ - 5}}$
The process of determining the $pH$ value of the solution is possible using the equation:
$pH = p{K_a} + \log \dfrac{{\left[ {Salt} \right]}}{{\left[ {Acid} \right]}}$
Here the $\dfrac{{\left[ {Salt} \right]}}{{\left[ {Acid} \right]}}$ is the ratio of the sodium salt with respect to the acetic acid.
In this case from ${K_a} = 1.8 \times {10^{ - 5}}$ the value of $p{K_a}$ can be determined. The value of $p{K_a}$ is the negative logarithm of dissociation constant value. The value of $p{K_a}$ can be calculated as:
\[p{K_a} = - \log \left( {1.8 \times {{10}^{ - 5}}} \right)\]
$ \Rightarrow p{K_a} = - \{ \log (1.8 \times {10^{ - 5}})\} $
$ \Rightarrow p{K_a} = - \{ \log (1.8) + \log ({10^{ - 5}})\} $ The logarithms are separated
$ \Rightarrow p{K_a} = - \{ \log \left( {1.8} \right) - 5\log (10)\} $ Power is brought as the coefficient
$ \Rightarrow p{K_a} = - \{ 0.255 - 5\} $ Putting the values of logarithm and log10=1 $ \Rightarrow p{K_a} = - \{ - 4.745\} $
Therefore, the value $p{K_a} = 4.745$
Hence from the given equation, putting the values of $pH$ given and the $p{K_a}$ calculated we get,
$5.7 = 4.745 + \log \dfrac{{\left[ {Salt} \right]}}{{\left[ {Acid} \right]}}$
This shows that, \[\log \dfrac{{\left[ {Salt} \right]}}{{\left[ {Acid} \right]}} = 5.7 - 4.745\]
Hence, we get the $\log \dfrac{{\left[ {Salt} \right]}}{{\left[ {Acid} \right]}} = 0.955$
Removing log on both sides we get, $\dfrac{{\left[ {Salt} \right]}}{{\left[ {Acid} \right]}} = 9.02$
In the given condition the ratio of the mixture required is $9.02$.
Note: The specific concentration of acid and the salt can be mixed in a solution to attain a particular $pH$ level. Thus, by adding a specific level of acid and salt a predetermined $pH$ can be attained. This is the importance of the specific components resulting in the particular $pH$.
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