In a 1.760 M aqueous solution of a monoprotic acid, 3.21% of the acid is ionized. What is the value of it's Ka?
x=1.760M * 3.21/100 - .0565M
Ka expression is Ka= {[H+][A-]} / [HA]
Did you make a typo; that = 0.0565M and not -.
All of that looks good to me.
To find the value of the Ka for the monoprotic acid in the given solution, we need to determine the concentrations of the different species involved in the equilibrium. We can start by calculating the concentration of the ionized acid, [A-], using the given information.
Step 1: Calculate the concentration of ionized acid [A-]:
To find the concentration of [A-], we can multiply the concentration of the original solution (1.760 M) with the percentage of acid ionized (3.21/100).
[A-] = 1.760 M * (3.21/100) = 0.0565 M
Step 2: Calculate the concentration of undissociated acid [HA]:
Since the given solution is a monoprotic acid and we know the concentration of the original solution (1.760 M), the concentration of [HA] will be the same as the original solution concentration.
[HA] = 1.760 M
Step 3: Calculate the concentration of hydronium ion [H+]:
To calculate the concentration of [H+], we can use the fact that for every molecule of ionized acid, we get an equal concentration of H+.
[H+] = [A-] = 0.0565 M
Step 4: Calculate the value of Ka:
Using the Ka expression: Ka = ([H+][A-])/[HA], we can substitute the values we found:
Ka = (0.0565 M * 0.0565 M) / 1.760 M
Ka = 0.001823 M
Therefore, the value of Ka for the monoprotic acid in the given solution is 0.001823 M.