Achieve Week 2 #6

[Jacqueline Vargas 3L]

Hi.

I'm having trouble approaching this question: "HClO is a weak acid (Ka=4.0×10^−8) and so the salt NaClO acts as a weak base. What is the pH of a solution that is 0.079 M in NaClO at 25 °C?"

The hint that is given is "Na+ is a spectator ion in this situation. Start by finding the Kb of ClO−."

An spectator ion is one that is both in the reactant and the product sides of the chemical equation, but I'm not sure how to use this information to find the Kb. Could someone clarify this for me?

Thank you in advance.

[905749989]

The formula you need to use is Ka x Kb = 1.0 x 10^-14. Then you would use and ice table to find the OH- concentration.

[Madison Chan 3B]

Hi!
In this problem, HClO is the weak acid and NaClO is the weak base. However, the salt (NaClO) has the cation Na+ which doesn't affect the pH of the solution so you can omit it in calculations for pH.

Then I wrote out the reaction and put the salt first: ClO (aq) + H2O (l) ⇌ HClO (aq) + OH- (aq)
Using an ICE table, I wrote out the equation for Kb using x-values. Since we are given a Ka value, we need to use Ka * Kb = Kw = 1.0 * 10^-14 to solve for Kb.
From there, you can solve for x, find the pOH, and then find the pH.

hope this helps!

[Genelle Marcelino-Searles 2G]

To find the pH, you need to use an ICE table to solve for Kb. To get this Kb, however, you need to use the equation that Ka x Kb=1x10^-14. Once you find this, you can solve for x and then find the pOH.

[Jacqueline Vargas 3L]

Hi!
In this problem, HClO is the weak acid and NaClO is the weak base. However, the salt (NaClO) has the cation Na+ which doesn't affect the pH of the solution so you can omit it in calculations for pH.

Then I wrote out the reaction and put the salt first: ClO (aq) + H2O (l) ⇌ HClO (aq) + OH- (aq)
Using an ICE table, I wrote out the equation for Kb using x-values. Since we are given a Ka value, we need to use Ka * Kb = Kw = 1.0 * 10^-14 to solve for Kb.
From there, you can solve for x, find the pOH, and then find the pH.

hope this helps!

This was super helpful! Thank you so much.

Another thing regarding this question. I was able to solve for x by assuming it was a small value, but I also tried using the quadratic equation. However, I ran into the problem of the value being undefined. I triple checked my work for the quadratic equation method and still had an undefined value. Did anyone else ran into a similar issue for this problem?