Question

Calculating the pH of a Weak Acid Calculate the pH of a 1 M solution of hydrofluoric acid (HF, K, 7.2 x 101)

Answer 1

Answer: The pH of weak acid is 1.001

Step-by-step explanation:

We are given:

Concentration of HF = 0.1 M

The chemical equation for the dissociation of HF follows:

`HF\rightarrow H^++F^-`

Initial: 0.1

At eqllm: 0.1-x x x

The equation for equilibrium constant follows:

`K_a=([H^+][F^-])/([HF])`

We are given:

`K_a=7.2* 10^1`

Putting values in above equation, we get:

`7.2* 10^1=(x* x)/((0.1-x))\\\\x=-72.09,0.0998`

Neglecting the negative value of 'x' because concentration cannot be negative

So, concentration of
`H^+` = 0.0998 M

To calculate the pH of the solution, we use the equation:

`pH=-\log[H^+]`

`pH=-\log(0.0998)`

`pH=1.001`

Hence, the pH of weak acid is 1.001

Answer 2

pH = 1.6

Step-by-step explanation:

• HF + H2O ↔ H3O+ + F-
• 2H2O ↔ H3O+ + OH-
• Ka = ( [ H3O+ ] * [ F- ] ) / [ HF ]

mass balance:

1 M = [ HF ] + [ F- ].........(1)

charge balance:

[ H3O+ ] = [ F- ] + [ OH- ]......[ OH- ] : comes from water, therefore it is negligible.

⇒ [ H3O+ ] = [ F- ]...........(2)

(2) in (1):

1 M = [ HF ] + [ H3O+ ]

⇒ [ HF ] = 1 - [ H3O+ ].........(3)

(2) and (3) in Ka:

⇒ Ka = [ H3O+ ]² / ( 1 - [ H3O+ ] )

∴ Ka = 7.0 E-4.......from literature

⇒ [ H3O+ ]² + 7.0 E-4[ H3O+ ] - 7.0 E-4 = 0

⇒ [ H3O+ ] = 0.0261 M

⇒ pH = 1.6