HW 2 Q 7/8

[Sophia Spallone 1A]

Hi, I'm having trouble on HW 2 questions 8 and 7. Specifically, I'm confused about how (for question 7) HClO and NaClO are related. I know that HClO is an acid and that NaClO is a base, and that their equations look like this:

HClO (aq) + H2O (l) ---> ClO- (aq) + H3O+ (aq)

NaClO (aq) + H2O (l) ---> NaClOH+ (aq) + OH- (aq)

We know the Ka of HClO, and the molar concentration of NaClO, and the question asks for the pH of the NaClO solution. Can someone help explain how you can connect these two reactions so that the Ka of HClO can be used to find the pH of the NaClO solution?

[Emily_Haywood_1K]

In this equation, Na is a spectator ion and therefore you can ignore it. It does not have any effect on the pH of the final solution. Therefore, to solve for the pH, use the equation ClO- + H20---> HClO + OH-. You can derive the Kb value from the Ka via the equation Ka+Kb= 10^-14. Once you have the Kb value, solve for the concentration of OH- ions using an ice table. Then use the equation pKa + pKb = 14 to solve for the pH.

[Elise Van Meter 2H]

Hi! I think that after you consider the first dissociation and you find that HClO (aq) + H2O (l) ---> ClO- (aq) + H3O+ (aq), you want to consider the further dissociation of the ClO- ion, which when combined with water will yield the following equation ClO- (aq) + H2O (l) ---> ClOH (aq) + OH- (aq). Then, as the stoichiometric ratio between the original HClO and the yielded ClO- is 1:1, you can use an ICE table to determine the resulting concentration of the OH- ion, such that the equilibrium concentrations are as follows: ClO-: original concentration - x, HClO: x, OH-: x. Then, given that Kb x Ka = Kw, you can solve for Kb to determine the value of x and thus the concentration of OH-. You can then solve for pOH (pOH = -log[OH-]), and then solve for pH (pH + pOH = 14). I hope this is helpful!

[Jessica_Lin_1J]

For both #7 and #8 on the homework, we have to first consider the relationship KA * KB = 1.0 * 10^-14.

Once you have found the respective KB or KA for #7 and #8, depending on what is given by the question, continue to do your ICE table like normal.

The additional elements (Na and Cl) will essentially be neutral and can be omitted when completing the ICE table and solving for X.

With X, you know now the equilibrium concentration of OH- or H+ (depending on which question you are solving)

Take the negative log of OH- to solve for pOH and solve for pH OR take the negative log of H+ to find the pH (again, depending on which question you are solving for).

Note: This question might initially throw you off because of the additional elements and/or KA and KB relationship. But they are essentially the same question with different values and start with either KA or KB.

[stella]

I believe that when NaClO dissociates it creates Na+ and ClO-, and ClO- is a conjugate of HClO because when HClO dissociates, it dissociates into H+ and ClO-. This means we can use HClO Ka they gave us to find the Kb of NaClO using Kw= Ka x Kb

[Krikor_Kevranian_1F]

For homework questions 7 and 8. After you get your x value and use -log to find the pOH you can simplify the pH+poH=14 where you can use subtraction to get the pH of the solution. Hope this helps!