Achieve question 7 and 8

[Skylar Smith 1E]

Hi! I have a quick question, how do we write out the equations for acids and bases when they give you a weak acid that turns into a weak base? We had to do this for problems 7 and

[Erin Nagel 1G]

When you are given the Ka for a weak acid and told to use its conjugate base in a reaction and calculate the pH for that reaction, you will need to convert the Ka given into the Kb for the conjugate base. To do this divide Kw (1.0 x 10^-14) by the Ka value. Then set up the equation the way you would for any acid reaction with HA (aq) + H2O (l) -> A- (aq) + H3O+ (aq).

[Matthew Morita 2F]

Hello,
So for setting up equations, you wanna start with what the question is asking for. So for example, question 8 is asking for you to find the pH of the NH4Cl. So you start the equation off with NH4+ (since CL- is neutral) and then the products should be NH3 and H3O+. This also makes sense because you use Ka for this problem and you start the equation off with an acid. Hope this makes sense

[Jess Max 2F]

When you are given the Ka for a weak acid and told to use its conjugate base in a reaction and calculate the pH for that reaction, you will need to convert the Ka given into the Kb for the conjugate base. To do this divide Kw (1.0 x 10^-14) by the Ka value. Then set up the equation the way you would for any acid reaction with HA (aq) + H2O (l) -> A- (aq) + H3O+ (aq).

Hi! This explanation is very helpful but I'm still tripped up on setting up the equilibrium table. I'm confused about which side of the equation the HClO and NaClO each go on. Are they both reactants? Do we assume H2O is also a reactant on the left?

[lanatruong]

Hi, so for question 7, NaClO dissociates into the neutral Na + ion and the basic ClO− ion. You can see that HClO and ClO− are conjugates, so the equation would look like
ClO- (aq) + H2O (l) -> HClO (aq) + OH- (aq) and you would disregard Na+ in the equation because it is neutral. Then you would solve for Kb from the given Ka value with the equation Kb=Kw/Ka. After doing this step, you can set up the ICE table with the given concentration of NaClO being in the initial column for ClO- and 0 for HClO and OH-. After doing this you can solve for X with the Kb value you found. Lastly, you can solve for pOH=-log[OH-] which is equal to pOH=-log(X). With the pOH value you can easily solve for pH=14-pOH to get your final answer. Hope this helps!

[May Jiang 1E]

The important thing to remember is that when they give you something where a weak acid turns into a weak base, it means that the weak acid has been added to a strong base. In 7, HClO turned into the weak base NaClO, so when we look at our list of strong bases, we find that adding NaOH to HClO can make NaClO, so NaOH + HClO makes NaClO and H2O. Hope this helps.

[Emmy_Adler_2K]

anyone know why we use the Kb value on #7 and not the Ka value? aren't we working with an acid?