Respuesta :
Answer : The pH of 0.289 M solution of lithium acetate at [tex]25^oC[/tex] is 9.1
Explanation :
First we have to calculate the value of [tex]K_b[/tex].
As we know that,
[tex]K_a\times K_b=K_w[/tex]
where,
[tex]K_a[/tex] = dissociation constant of an acid = [tex]1.8\times 10^{-5}[/tex]
[tex]K_b[/tex] = dissociation constant of a base = ?
[tex]K_w[/tex] = dissociation constant of water = [tex]1\times 10^{-14}[/tex]
Now put all the given values in the above expression, we get the dissociation constant of a base.
[tex]1.8\times 10^{-5}\times K_b=1\times 10^{-14}[/tex]
[tex]K_b=5.5\times 10^{-10}[/tex]
Now we have to calculate the concentration of hydroxide ion.
Formula used :
[tex][OH^-]=(K_b\times C)^{\frac{1}{2}}[/tex]
where,
C is the concentration of solution.
Now put all the given values in this formula, we get:
[tex][OH^-]=(5.5\times 10^{-10}\times 0.289)^{\frac{1}{2}}[/tex]
[tex][OH^-]=1.3\times 10^{-5}M[/tex]
Now we have to calculate the pOH.
[tex]pOH=-\log [OH^-][/tex]
[tex]pOH=-\log (1.3\times 10^{-5})[/tex]
[tex]pOH=4.9[/tex]
Now we have to calculate the pH.
[tex]pH+pOH=14\\\\pH=14-pOH\\\\pH=14-4.9=9.1[/tex]
Therefore, the pH of 0.289 M solution of lithium acetate at [tex]25^oC[/tex] is 9.1
