Question
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
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GPT 3.5
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.
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.
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