The weak acid HQ has pKa of 4.89. Calculate the [H3O+] of .035 M HQ.
Thanks.
..........HQ ==> H^+ + Q^-
I......0.035.....0.....0
C........-x......x.....x
E.......0.035-x..x.....x
Convert pKa to Ka, then substitute the E line into Ka expression and solve for x. Post your work if you get stuck.
To calculate the concentration of H3O+ in a solution of a weak acid, you can use the dissociation constant (Ka) and the initial concentration of the acid.
The dissociation of the weak acid HQ can be represented as follows:
HQ ⇌ H+ + Q-
The equilibrium expression for this dissociation is:
Ka = [H+][Q-] / [HQ]
Given that HQ has a pKa of 4.89, we can use the formula pKa = -log10(Ka) to calculate the value of Ka:
pKa = -log10(Ka)
4.89 = -log10(Ka)
Now, solve for Ka by taking the antilog of both sides:
Ka = 10^(-4.89)
Next, let's assume that the concentration of HQ that you provided (0.035 M) is the initial concentration before any dissociation occurs.
Let x represent the concentration of [H+] that we want to find.
[H+] = x
[Q-] = x (assuming for each one dissociated, an equal amount of Q- is formed)
[HQ] = 0.035 M - x (since it dissociates into x amount of [H+] and x amount of [Q-])
Plugging the values into the equilibrium expression:
Ka = [H+][Q-] / [HQ]
10^(-4.89) = [x][x] / (0.035 - x)
Now, we can solve this equation to find x, which represents the concentration of [H+]. Let's solve it:
10^(-4.89) = x^2 / (0.035 - x)
Cross-multiply:
0.035 - x = x^2 * 10^(-4.89)
Rearrange the equation:
x^2 * 10^(-4.89) + x - 0.035 = 0
Now, we can solve this quadratic equation to find the value of x using numerical methods like the quadratic formula or a calculator. Solving for x gives us the concentration of [H+], which is the same as [H3O+].
Note: It's important to remember that the quadratic equation is necessary in this case because the dissociation of a weak acid is not complete, meaning it does not go to 100% completion.