Respuesta :
Answer : The pH of the solution is, 8.38
Solution : Given,
Concentration (c) = 0.11 M
Acid dissociation constant = [tex]k_a=1.8\times 10^{-4}[/tex]
First we have to calculate the moles of sodium formate.
[tex]Moles=\frac{Mass}{\text{Molar mass}}=\frac{0.680g}{68g/mole}=0.010mol[/tex]
Now we have to calculate the concentration of formic acid.
Concentration of formic acid = Concentration of sodium formate = [tex]\frac{Moles}{Volume}=\frac{0.010mol}{0.100L}=0.1M[/tex]
Now we have to calculate the base dissociation constant.
[tex]k_w=k_a\times k_b\\\\k_b=\frac{k_w}{k_a}\\\\k_b=\frac{1\times 10^{-14}}{1.8\times 10^{-4}}\\\\k_b=5.6\times 10^{-11}[/tex]
The equilibrium reaction for dissociation of (weak base) is,
[tex]HCOO^-+H_2O\rightleftharpoons HCOOH+OH^-[/tex]
initially conc. 0.1 0 0
At eqm. (0.1-x) x x
First we have to calculate the value of 'x'.
Formula used :
[tex]k_b=\frac{[HCOOH][OH^-]}{[HCOO^-]}[/tex]
Now put all the given values in this formula ,we get:
[tex]5.6\times 10^{-11}=\frac{(x)(x)}{(0.1-x)}[/tex]
By solving the terms, we get
[tex]x=2.4\times 10^{-6}M[/tex]
Thus, the concentration of hydroxide ion is:
[tex][OH^-]=x=2.4\times 10^{-6}M[/tex]
Now we have to calculate the pOH.
[tex]pOH=-\log [OH^-][/tex]
[tex]pOH=-\log (2.4\times 10^{-6})[/tex]
[tex]pOH=5.62[/tex]
Now we have to calculate the pH.
[tex]pH+pOH=14\\\\pH=14-pOH\\\\pH=14-5.62\\\\pH=8.38[/tex]
Therefore, the pH of the solution is, 8.38
