A certain monoprotic weak acid with Ka = 6.7 × 10-8 can be used in various industrial processes, but is no longer used in chemistry teaching labs because it tends to explode when it gets old. (a) What is the [H+] for a 0.558 M aqueous solution of this acid and (b) what is its pH?
PLEASE HELP!!
...........HA ==> H^+ + A^-
begin....0.558M...0.....0
change.....-x......x.....x
final...0.558-x....x.....x
Ka = (H^+))A^-)/(HA)
Substitute the ICE chart values I made into the Ka expession and solve for H^+.
b) pH = -log(H^+)
Thank you so much!!
I HAVE NO F$#@ING IDEA!!!!!!!!!!!!!
i need help with worksheet Titration worksheet I
To find the [H+] for the given concentration of the weak acid, we can use the equilibrium expression for the ionization of the acid:
HA ⇌ H+ + A-
The equilibrium constant (Ka) for this reaction is given as 6.7 × 10^-8.
(a) To determine the concentration of [H+], we need to find the equilibrium concentration of the acid (HA) that has ionized. Let's assume that 'x' is the equilibrium concentration of [H+]. Therefore, the concentration of [A-] (conjugate base) will also be 'x' since the ratio of [H+] to [A-] is 1:1.
The initial concentration of the acid (HA) is 0.558 M, and since the concentration of [H+] and [A-] will be the same, we can substitute 'x' for the concentration of both [H+] and [A-] in the equilibrium expression:
Ka = [H+][A-]/[HA]
6.7 × 10^-8 = x * x / 0.558
Simplifying the equation:
x^2 = (6.7 × 10^-8) * 0.558
x^2 = 3.7486 × 10^-8
Taking the square root of both sides:
x = √(3.7486 × 10^-8)
x = 6.12 × 10^-4
Therefore, the [H+] concentration for a 0.558 M aqueous solution of the weak acid is 6.12 × 10^-4 M.
(b) To find the pH of the solution, we can use the equation:
pH = -log[H+]
Substituting the [H+] concentration we found above:
pH = -log(6.12 × 10^-4)
Calculating the pH:
pH ≈ 3.21
Therefore, the pH of the solution is approximately 3.21.