
The ${{K}_{a}} $ of propionic acid is $1.34\times {{10} ^ {-5}} $. What is the pH of a solution containing 0.5M propionic acid and 0.5M sodium propionate?
Answer
474.9k+ views
Hint: For calculating pH we need to find:
$p{{K}_{a}} $, by using the formula: $p{{K}_{a}} =-\log {{K}_{a}} $,
Then, after finding $p{{K}_{a}} $, pH can be calculated by: $pH=p{{K}_{a}} +\log \left (\dfrac {\left [ salt \right]} {\left [ acid \right]} \right) $.
Complete step by step solution:
We have been provided that ${{K}_{a}} $ of propionic acid is $1.34\times {{10} ^ {-5}} $,
An acid dissociation constant (${{K}_{a}} $) is a quantitative measure of the strength of an acid in solution. It is the equilibrium constant for a chemical reaction known as dissociation in the context of acid–base reactions.
We know that $p{{K}_{a}} =-\log {{K}_{a}} $,
Now, keeping the values in the above equation we would get: $p{{K}_{a}} =-\log (1.34\times {{10} ^ {-5}}) $,
Solving it further: $p{{K}_{a}} =5-0.127$,
Value of $p{{K}_{a}} $ comes out to be: $p{{K}_{a}} =4.873$,
Now, we need to calculate the pH,
We know that pH is a scale used to specify the acidity or basicity of an aqueous solution. Acidic solutions are measured to have lower pH values than basic or alkaline solutions. The pH scale is logarithmic and inversely indicates the concentration of hydrogen ions in the solution.
So, we will be calculating the value of pH by using the formula: $pH=p{{K}_{a}} +\log \left (\dfrac {\left [ salt \right]} {\left [ acid \right]} \right) $,
Now, as we have been given the value of: $\left [ salt \right] =\left [ acid \right] =0.5M$,
Keeping these values in $pH=p{{K}_{a}} +\log \left (\dfrac {\left [ salt \right]} {\left [ acid \right]} \right) $,
We would get: $pH=4.87+\log \left (\dfrac {0.5} {0.5} \right) $,
So, the value of pH comes out to be: 4.873.
Note: pH is important because substances such as our stomach acids tend to be at a certain pH in order to work properly. pH is also important because it must be at certain levels in order for living organisms to survive.
$p{{K}_{a}} $, by using the formula: $p{{K}_{a}} =-\log {{K}_{a}} $,
Then, after finding $p{{K}_{a}} $, pH can be calculated by: $pH=p{{K}_{a}} +\log \left (\dfrac {\left [ salt \right]} {\left [ acid \right]} \right) $.
Complete step by step solution:
We have been provided that ${{K}_{a}} $ of propionic acid is $1.34\times {{10} ^ {-5}} $,
An acid dissociation constant (${{K}_{a}} $) is a quantitative measure of the strength of an acid in solution. It is the equilibrium constant for a chemical reaction known as dissociation in the context of acid–base reactions.
We know that $p{{K}_{a}} =-\log {{K}_{a}} $,
Now, keeping the values in the above equation we would get: $p{{K}_{a}} =-\log (1.34\times {{10} ^ {-5}}) $,
Solving it further: $p{{K}_{a}} =5-0.127$,
Value of $p{{K}_{a}} $ comes out to be: $p{{K}_{a}} =4.873$,
Now, we need to calculate the pH,
We know that pH is a scale used to specify the acidity or basicity of an aqueous solution. Acidic solutions are measured to have lower pH values than basic or alkaline solutions. The pH scale is logarithmic and inversely indicates the concentration of hydrogen ions in the solution.
So, we will be calculating the value of pH by using the formula: $pH=p{{K}_{a}} +\log \left (\dfrac {\left [ salt \right]} {\left [ acid \right]} \right) $,
Now, as we have been given the value of: $\left [ salt \right] =\left [ acid \right] =0.5M$,
Keeping these values in $pH=p{{K}_{a}} +\log \left (\dfrac {\left [ salt \right]} {\left [ acid \right]} \right) $,
We would get: $pH=4.87+\log \left (\dfrac {0.5} {0.5} \right) $,
So, the value of pH comes out to be: 4.873.
Note: pH is important because substances such as our stomach acids tend to be at a certain pH in order to work properly. pH is also important because it must be at certain levels in order for living organisms to survive.
Recently Updated Pages
Master Class 11 Accountancy: Engaging Questions & Answers for Success

Glucose when reduced with HI and red Phosphorus gives class 11 chemistry CBSE

The highest possible oxidation states of Uranium and class 11 chemistry CBSE

Find the value of x if the mode of the following data class 11 maths CBSE

Which of the following can be used in the Friedel Crafts class 11 chemistry CBSE

A sphere of mass 40 kg is attracted by a second sphere class 11 physics CBSE

Trending doubts
10 examples of friction in our daily life

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

State and prove Bernoullis theorem class 11 physics CBSE

What organs are located on the left side of your body class 11 biology CBSE

Write down 5 differences between Ntype and Ptype s class 11 physics CBSE
