Question

Question 14 (5 points)

What's the acid ionization constant for an acid with a pH of 2.11 and an equilibrium
concentration of 0.30 M?
O A) 4.87x10-8
B) 1.99x10-6
C) 3.32x10-4
OD) 2.01x10-4

Answer 1

D) 2.01 x 10⁻⁴ is correct!

Step-by-step explanation:

I got it in class!

Hope this Helps!! :))

Answer 2

D) 2.01 x 10⁻⁴ .

Step-by-step explanation:

pH = 2.11

[ H⁺ ] =
`10^(-2.11)`

Let the acid be HA

It will ionise as follows .

HA ⇄ H⁺ + A⁻

in equilibrium .30
`10^(-2.11)`
`10^(-2.11)`

Acid ionisation constant Ka =
`((10^(-2.11))^2)/(0.3)`

= 2 x 10⁻⁴