
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
561.3k+ views
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 $pH$.
Recently Updated Pages
Master Class 11 Accountancy: Engaging Questions & Answers for Success

Master Class 11 Science: Engaging Questions & Answers for Success

Master Class 11 Business Studies: Engaging Questions & Answers for Success

Master Class 11 English: Engaging Questions & Answers for Success

Master Class 11 Computer Science: Engaging Questions & Answers for Success

Master Class 9 General Knowledge: Engaging Questions & Answers for Success

Trending doubts
Which are the Top 10 Largest Countries of the World?

What are the major means of transport Explain each class 12 social science CBSE

Differentiate between insitu conservation and exsitu class 12 biology CBSE

Draw a labelled sketch of the human eye class 12 physics CBSE

State the principle of an ac generator and explain class 12 physics CBSE

Give 10 examples of unisexual and bisexual flowers

