Question

What is the $pH$ of a $0.150M$ solution of sodium acetate ( $NaO_2CCH_3$)? $Ka(CH_3CO_2H) = 1.8 \times 10^ { -5 }.$

Answer 1

For solving the given problem, we should have knowledge about salt hydrolysis, $pH$, acidic constant $Ka$ and calculation of $log$.
Salt hydrolysis is a process where the reactants are salt and water forming an acidic or basic solution. $pH$ is described as power of hydrogen or potential of hydrogen. It is used to describe the concentration of hydrogen and is inversely proportional to it.
Acidic constant or Acid Dissociation constant, $Ka$ is used to quantitatively measure the acidic strength in a solution.
$pH$ for basic solution;
$pH = 7+\dfrac{ 1 }{ 2 } [P_{ Ka } + logC ]$
Where, C is concentration of salt.
$P_{ Ka }$ can be calculated as ;
$P_{ Ka } = -log Ka$.

Complete step by step solution:
Step-1 :
First we have to determine that the given salt is if hydrolysed contains what products. On hydrolysis of Sodium acetate, $NaO_2CCH_3$, we get a strong base $NaOH$ and a weak acid $(CH_3CO_2H)$ (acetic acid).
Step-2 :
The formula for $pH$ of salt of strong base and weak acid is :
$pH = 7+\dfrac{ 1 }{ 2 } [P_{ Ka } + logC ]$
Since, the $pH$ is basic.
Step-3 :
Here, we have given $Ka = 1.8 \times 10^{ -5 }$ and $C = 0.15M$.
For, $P_{ Ka } = -log Ka$
$=-log(1.8 \times 10^{ -5 }$
$=4.76$
Putting the above values in formula, we get;
$pH = 7+\dfrac{ 1 }{ 2 } [4.76 + 0.15 ]$
$=9.455$

Note:
While calculating the $pH$, make sure that salt hydrolysis is correct and the $pH$ is according to that acidic or basic solution, for basic solution, $pH$ is added and for acidic solution, $pH$ is subtracted.