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:

pH = 1.6

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

Answer 2

Answer: The pH of weak acid is 1.001

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:

$K_a=\frac{[H^+][F^-]}{[HF]}$

We are given:

$K_a=7.2\times 10^1$

Putting values in above equation, we get:

$7.2\times 10^1=\frac{x\times x}{(0.1-x)}\\\\x=-72.09,0.0998$

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