Question

Lithium acetate, LiCH3CO2, is a salt formed from the neutralization of the weak acid acetic acid, CH3CO2H, with the strong base lithium hydroxide. Given that the value of Ka for acetic acid is 1.8×10−5, what is the pH of a 0.289 M solution of lithium acetate at 25∘C?

Answer 1

Answer : The pH of 0.289 M solution of lithium acetate at $25^oC$ is 9.1

Explanation :

First we have to calculate the value of $K_b$.

As we know that,

$K_a\times K_b=K_w$

where,

$K_a$ = dissociation constant of an acid = $1.8\times 10^{-5}$

$K_b$ = dissociation constant of a base = ?

$K_w$ = dissociation constant of water = $1\times 10^{-14}$

Now put all the given values in the above expression, we get the dissociation constant of a base.

$1.8\times 10^{-5}\times K_b=1\times 10^{-14}$

$K_b=5.5\times 10^{-10}$

Now we have to calculate the concentration of hydroxide ion.

Formula used :

$[OH^-]=(K_b\times C)^{\frac{1}{2}}$

where,

C is the concentration of solution.

Now put all the given values in this formula, we get:

$[OH^-]=(5.5\times 10^{-10}\times 0.289)^{\frac{1}{2}}$

$[OH^-]=1.3\times 10^{-5}M$

Now we have to calculate the pOH.

$pOH=-\log [OH^-]$

$pOH=-\log (1.3\times 10^{-5})$

$pOH=4.9$

Now we have to calculate the pH.

$pH+pOH=14\\\\pH=14-pOH\\\\pH=14-4.9=9.1$

Therefore, the pH of 0.289 M solution of lithium acetate at $25^oC$ is 9.1