Calculate the value of [H3O+] in a 0.01 M HOBr solution. Ka = 2.5E-9
I'm having a problem with just writing the equilibrium expression.
do you add H2O to the HOBr?
HOBr + H2O <---> H3O+ + OBr-
but then you could say that because HOBr = .01 M, then 0Br- = .01 M, and H3O+ would equal the Ka. but that is not the answer...
Yes, technically, one adds water but many people omit the water.
HOBr + H2O ==> H3O^+ + OBr^- OR
HOBr ==> H^+ + OBr^-
Ka = 2.5E-9 = (H3O^+)(OBr^-)/(HOBr)
What you are doing wrong is you are assuming the ionization is 100% and it isn't (otherwise the Ka would be infinity). A solution 0.01 M in HCl will be 0.01M in H^+ and 0.01M in Cl^-; but that is because HCl is 100% ionized. HOBr, having a Ka of 2.5E-9 isn't anywhere close to 100% and you cannot do it this way. You set up an ICE chart, substitute into Ka expression, and solve for the unknown.
initial:
HOBr = 0.01
H^+ = 0
OBr^- = 0
change:
H^+ = +x
OBr^- = +x
HOBr = -x
equilibrium:
H^+ = +x
OBr^- = +x
HOBr = 0.01-x
Substitute into Ka expression above and solve for x.
thanks!!!
The equilibrium expression you provided is correct:
HOBr + H2O ⇌ H3O+ + OBr-
When writing the equilibrium expression, water (H2O) is not included because it is considered a solvent in this case. Only species that are involved in the chemical reaction are included.
To solve the problem, you need to set up an ICE (Initial, Change, Equilibrium) table. Let's assume x represents the concentration of H3O+ formed at equilibrium.
The initial concentrations are:
[HOBr] = 0.01 M
[H2O] = not included in the equilibrium expression
[H3O+] = 0 M (because it hasn't formed yet)
[OBr-] = 0 M (because it hasn't formed yet)
The changes in concentration are:
[HOBr] - x
[H2O] - x
[H3O+] + x
[OBr-] + x
The equilibrium concentrations are:
[HOBr]eq = 0.01 - x
[H2O]eq = not included in the equilibrium expression
[H3O+]eq = x
[OBr-]eq = x
Using the given Ka value, you can write the equilibrium expression as:
Ka = ([H3O+]eq * [OBr-]eq) / [HOBr]eq
Substituting the equilibrium concentrations:
2.5E-9 = (x * x) / (0.01 - x)
Since the value of x is expected to be small compared to 0.01, you can approximate 0.01 - x as 0.01 and rewrite the equation as:
2.5E-9 = x^2 / (0.01)
Rearranging the equation, you get:
2.5E-9 * 0.01 = x^2
x^2 = 2.5E-11
Taking the square root,
x ≈ 5E-6
Therefore, the concentration of [H3O+] in the 0.01 M HOBr solution is approximately 5E-6 M.
To calculate the value of [H3O+] in a 0.01 M HOBr solution, you need to consider the equilibrium expression and use the given Ka value. Here's how you can approach it:
Step 1: Write the balanced equation for the dissociation of HOBr:
HOBr + H2O ⇌ H3O+ + OBr-
Step 2: Write the equilibrium expression using the concentrations of the species involved:
Ka = [H3O+][OBr-]/[HOBr]
Step 3: Determine the concentrations of the species.
In this case, you are given that the concentration of HOBr is 0.01 M. However, you cannot assume that the concentration of OBr- is also 0.01 M because they are not equal. The equilibrium expression accounts for the fact that the concentrations can be different.
Step 4: Solve for the unknown concentration [H3O+].
Rearrange the equilibrium expression to solve for [H3O+]:
[H3O+] = (Ka * [HOBr])/[OBr-]
Step 5: Substitute the given values into the equation.
[H3O+] = (2.5E-9 * 0.01 M)/[OBr-]
Since you haven't provided the concentration of OBr-, we cannot calculate the exact value of [H3O+] without that information. The equilibrium constant Ka only tells us the ratio of products to reactants at equilibrium.
To calculate the value of [H3O+], you would need to know either the initial concentration of OBr- or have additional information that allows you to determine its concentration at equilibrium, such as a second equilibrium expression or additional data.