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Sedbober [7]
2 years ago
13

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?
Chemistry
1 answer:
Vesna [10]2 years ago
4 0

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

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