1- Calculate the equilibrium concentration for H+, HSO4- and SO4^2- , given a .150M H2SO4 solution. the Ka for HSO4- is 1.20*10^-2
The first ionization is 100%; therefore,
..........H2SO4 ==> H^+ + HSO4^-
I.........0.150......0.....0
C........-0.l50...0.150..0.150
The second H is weak.
............HSO4^- ==> H^+ + SO4^2-
I.........0.150......0.150..0
C...........-x.........+x...x
E........0.150-x....0.150+x..x
k2 = 0.012 = (H^+)(SO4^2-)/(HSO4^-)
Substitute the E line of the second ICE chart and solve for x, then
total (H^+) = 0.150+x
(SO4^2-) = x
(HSO4^-) = 0.150-x
Of course (H2SO4) = 0
Ah, acid-base equilibrium. Time to put on my chemistry hat! But first, let me entertain you with a joke: Why can't you trust atoms? Because they make up everything!
Now, let's focus on answering your question. We know that H2SO4 dissociates into H+ and HSO4- in water. And, since the Ka for HSO4- is given, we can use that information to calculate the equilibrium concentrations.
Let's denote the equilibrium concentration for H+ as [H+], HSO4- as [HSO4-], and SO4^2- as [SO4^2-]. Since H2SO4 is a strong acid, it completely dissociates, meaning the initial concentration of [H+] is equal to the concentration of the H2SO4 solution (0.150 M).
To calculate the equilibrium concentration of HSO4- and SO4^2-, we'll use the Ka expression for the dissociation of HSO4-:
Ka = [H+][SO4^2-] / [HSO4-]
Given Ka as 1.20*10^-2 and the initial concentration of [H+] as 0.150 M, we can rearrange the equation to solve for [HSO4-]:
0.150 * [SO4^2-] / [HSO4-] = 1.20*10^-2
Now, let the equilibrium concentration of HSO4- be represented as "x". Thus:
0.150 * [SO4^2-] / x = 1.20*10^-2
Now we can solve for x, which represents the equilibrium concentration of HSO4-:
x = (0.150 * [SO4^2-]) / (1.20*10^-2)
Substituting the Ka for HSO4-:
x = (0.150 * [SO4^2-]) / (1.20*10^-2)
To find the equilibrium concentration of SO4^2-, we need to consider the stoichiometry. Since H2SO4 dissociates into one HSO4- and one SO4^2-, the concentration of [SO4^2-] will be the same as [HSO4-]. Therefore, the equilibrium concentration of [SO4^2-] will also be "x".
So, the equilibrium concentrations are:
[H+] = 0.150 M
[HSO4-] = x
[SO4^2-] = x
Now, go ahead and calculate the value of "x" using the equation provided, and you'll have your equilibrium concentrations!
To calculate the equilibrium concentration of H+, HSO4-, and SO4^2- in a .150M H2SO4 solution, we need to consider the dissociation reaction of H2SO4. The dissociation reaction is as follows:
H2SO4 ⇌ H+ + HSO4-
Based on the reaction equation, the initial concentration of H2SO4 is 0.150M. Let's assume that the equilibrium concentration of HSO4- is x. As H2SO4 is a strong acid, it completely dissociates into H+ and HSO4-, so the equilibrium concentration of H+ is also x.
Now, using the given Ka value of HSO4- (1.20*10^-2), we can set up an expression for the equilibrium constant, Ka:
Ka = [H+][HSO4-] / [H2SO4]
Since the concentration of H+ and HSO4- is equal to x, and the concentration of H2SO4 is 0.150M, we can substitute these values into the equilibrium constant expression:
1.20*10^-2 = x^2 / 0.150
Now, let's solve for x by rearranging the equation:
x^2 = 1.20*10^-2 * 0.150
x^2 = 1.8*10^-3
x = √(1.8*10^-3)
x ≈ 0.042
Thus, the equilibrium concentration of H+, HSO4-, and SO4^2- in the .150M H2SO4 solution is approximately 0.042M.
To calculate the equilibrium concentrations of ions in a solution of H2SO4, we need to use the acid dissociation constant (Ka) for the dissociation of HSO4-. The given value for Ka is 1.20 * 10^-2.
First, let's write the balanced chemical equation for the dissociation of H2SO4:
H2SO4 ⇌ H+ + HSO4-
Let's assume the initial concentration of H2SO4 is 0.150 M. At equilibrium, let's assume x moles per liter of H+ and HSO4- are formed.
Using the stoichiometry of the balanced equation, we can say that the concentration of H+ at equilibrium is also x M.
Since the concentration of HSO4- at equilibrium is equal to the initial concentration minus the concentration of H+ (0.150 - x), we can write:
Ka = [H+][HSO4-] / [H2SO4]
1.20 * 10^-2 = x * (0.150 - x) / 0.150
To solve this quadratic equation, we can rearrange it and form a quadratic equation:
x^2 - 0.150x + (0.150 * 1.20 * 10^-2) = 0
You can solve this equation using the quadratic formula or a calculator. Upon solving, you will get two possible values for x (let's denote them as x1 and x2). Since x represents the concentration, we discard any negative values as they are not physically meaningful for concentration.
Once we obtain the value of x, we can calculate the equilibrium concentrations as follows:
[H+] = x (M)
[HSO4-] = 0.150 - x (M)
[SO4^2-] = 0 (since H2SO4 does not dissociate into SO4^2-)
Remember to consider the significant figures and units in your final answer.