Achieve #1 week 2

[Tianna Nguyen 1I]

Hi, I'm not sure where to start for this problem: "If the Ka of a monoprotic weak acid is 8.1×10^-6, what is the pH of a 0.45 M solution of this acid?"

[Lauren Woodward 1I]

For this problem, use the basic form: HA -> H30+ + A-
First set up an ice table with your given concentrations. You are given the initial concentration of the weak acid, and the other initial values should be zero. At equilibrium, x amount of concentration will be subtracted from HA, and x will be added to each [H30+] and [A-], which were initially zero. Then set up the equation for Ka and solve for x.

[Omar Alami 3H]

The first thing that you should start out doing is writing the general formula for a weak acid which is HA-->(H3O+) + (A-). Then you should make an ice table for this equation. The initial concentration of the weak acid is given to you in the question as 0.45 M; the initial concentrations for the products would be 0 as they aren't given. Next, you fill in the values for the change row. The change for the reactant would -x as you are using up the initial concentration, but don't know by how much. This would make the change for both the products be +x and +x. Then combine the columns and that will get your equilibrium expressions that you will set equal to Ka given to you in the problem. Solve for x and take the -log of it to get you the value for the pH.

[Rio Gagnon 1G]

For this type of problem, you should be using an ice table. The general formula of the equation is acid --> H+ + conjugate base. The first row of the ice table should include the given molarity for the acid, along with 0 M for the 2 products. For the change row, subtract x from the acid and add x to each of the products. After your ice table is set up, solve for x by setting the equation equal to the given value for Ka. Since x is equal to H+, plug that value into the formula for pH (-log[H+]) to obtain your answer.

[Kitana_Garcia_3A]

For this problem, could we make the assumption that the value of X is smaller than the value given, so the denominator of our equation would just be the value given? For example, if it is x^2/0.42-x, would it be x^2/0.42?

[Adele Nguyen 2G]

For this problem, could we make the assumption that the value of X is smaller than the value given, so the denominator of our equation would just be the value given? For example, if it is x^2/0.42-x, would it be x^2/0.42?

Yes we can! The initial concentration (0.42 M) is a lot greater than the Ka given, meaning that the concentration change x is small.

[Melody Khoshneviszadeh 3C]

This problem definitely needs an ICE box set up as the measurements are done in relation to 0.45M and we need to find it the answer at equilibrium to measure the PH. Also important to note that monoprotic indicates that the acid only gives up one proton.

[Caroline 2A]

Hi!

First, start by assembling an ICE table. Because the value of M is more than 1000 times greater than the Ka, we are able to assume that the concentration change x is sufficiently small compared to the initial concentration. We can check this by dividing the value of X by the initial concentration and multiplying that value by 100%. If x is less than 5% of the concentration, then our assumption is valid (and it is valid in this case).

From there, we can find the pH by using the equation pH= -log[H+] where x is equal to H+. This should give you your final answer.

Hope this helps :)