Sapling #7 Week 2

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Rachel Kwan 1B
Posts: 40
Joined: Wed Nov 18, 2020 12:23 am

Sapling #7 Week 2

Postby Rachel Kwan 1B » Sun Jan 17, 2021 11:25 pm

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

Postby Manseej Khatri 2B » Sun Jan 17, 2021 11:42 pm

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 . The sodium cancel leaving the resulting equation.

jeffrey van 2I
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Re: Sapling #7 Week 2

Postby jeffrey van 2I » Mon Jan 18, 2021 9:42 am

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
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Re: Sapling #7 Week 2

Postby Kareena Patel 1G » Mon Jan 18, 2021 10:17 am

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: 33
Joined: Wed Nov 25, 2020 12:21 am

Re: Sapling #7 Week 2

Postby jeffrey van 2I » Mon Jan 18, 2021 10:53 am

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

Joseph Hsing 2C
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Re: Sapling #7 Week 2

Postby Joseph Hsing 2C » Mon Jan 18, 2021 12:31 pm

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.


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