## Sapling #7 Week 2

Rachel Kwan 1B
Posts: 47
Joined: Wed Nov 18, 2020 12:23 am

### Sapling #7 Week 2

For context here is the 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 0.045 M in NaClO at 25 °C?

My specific question is why do you use the equation Cl0- + H20 [equilibrium arrows] HCl0 + OH-? Why don't you use the equation involving NaClO?

Manseej Khatri 2B
Posts: 76
Joined: Wed Sep 30, 2020 9:42 pm

### Re: Sapling #7 Week 2

In solution Na+ and ClO- disassociate, with ClO- being the weak base (the conjugate of HClO4). The sodium does not participate in the reaction. The reaction could thus be represented as $Na^{+} + ClO^{^{-}} \Leftrightarrow Na^{+} + HClO + OH^{^{-}}$ . The sodium cancel leaving the resulting equation.

jeffrey van 2I
Posts: 40
Joined: Wed Nov 25, 2020 12:21 am

### Re: Sapling #7 Week 2

Would we use the Ka value given to find the pH, or do we need to convert the Ka value to Kb since the equation involves a base?

Kareena Patel 1G
Posts: 97
Joined: Wed Sep 30, 2020 9:58 pm
Been upvoted: 1 time

### Re: Sapling #7 Week 2

jeffrey van 2I wrote:Would we use the Ka value given to find the pH, or do we need to convert the Ka value to Kb since the equation involves a base?

You would need to convert it to Kb because it involves a base.

jeffrey van 2I
Posts: 40
Joined: Wed Nov 25, 2020 12:21 am

### Re: Sapling #7 Week 2

Thanks, so just to confirm, I would just divide the value of Kw (1.0E-14) by the given Ka, right?

Joseph Hsing 2C
Posts: 91
Joined: Wed Sep 30, 2020 9:42 pm

### Re: Sapling #7 Week 2

jeffrey van 2I wrote:Thanks, so just to confirm, I would just divide the value of Kw (1.0E-14) by the given Ka, right?

Yup this is correct in order to find Kb.