An equilibrium concentration of the H3O^+(aq) is 8.5 x 10^-4 mol/L. If the inital concentration of anthocyanins, prior to equilibrium was 1.62 x 10^-3 mol/L, the Ka for anthcyanins is
To find the value of Ka for anthocyanins, we need to set up the equation for the dissociation of anthocyanins.
The dissociation of anthocyanins can be written as:
HA ⇌ H+ + A-
Where HA represents anthocyanins in its undissociated form, H+ represents hydronium ions, and A- represents the conjugate base of anthocyanins.
The equilibrium constant expression, Ka, for this dissociation is given by:
Ka = [H+][A-]/[HA]
We are given the equilibrium concentration of H+ as 8.5 x 10^-4 mol/L. We can assume the concentration of A- is also 8.5 x 10^-4 mol/L since the dissociation is 1:1.
The initial concentration of HA is given as 1.62 x 10^-3 mol/L.
Now we can substitute the values into the Ka expression:
Ka = (8.5 x 10^-4 mol/L) * (8.5 x 10^-4 mol/L) / (1.62 x 10^-3 mol/L)
Ka ≈ 3.57 x 10^-7 mol/L
Therefore, the Ka for anthocyanins is approximately 3.57 x 10^-7 mol/L.