Question

A weak acid. What is the pH of a 0.1 M solution of acetic acid (pKa = 4.75)?

(Hint: Let x be the concentration of H+ ions released from acetic acid when it dissociates. The solutions to a quadratic equation of the form ax^2 + bx + c = 0 are x = (-b +- squareroot (b^2- 4ac)/2a.

Answer 1

pH of acetic acid solution is 2.88

Step-by-step explanation:

`pK_(a)=4.75`

or,
`-log(K_(a))=4.75`

or,
`K_(a)=10^(-4.75)=1.78* 10^(-5)`

We have to construct an ICE table to determine concentration of
`H^(+)` and corresponding pH. Initial concentration of acetic acid is 0.1 M.

`CH_(3)COOH\rightleftharpoons CH_(3)COO^(-)+H^(+)`

I(M): 0.1 0 0

C(M): -x +x +x

E(M): 0.1-x x x

So,
`([CH_(3)COO^(-)][H^(+)])/([CH_(3)COOH])=K_(a)`

or,
`(x^(2))/(0.1-x)=1.78* 10^(-5)`

or,
`x^(2)+(1.78* 10^(-5)* x)-(1.78* 10^(-6))=0`

So,
`x=\frac{-(1.78* 10^(-5))+\sqrt{(1.78* 10^(-5))^(2)+(4* 1* 1.78* 10^(-6))}}{2* 1}`(M)

so,
`x=1.33* 10^(-3)M`

Hence
`pH=-log[H^(+)]=-log(1.33* 10^(-3))=2.88`