6D.3a

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JamieVu_2C
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Joined: Thu Jul 25, 2019 12:16 am

6D.3a

Postby JamieVu_2C » Sun Jan 19, 2020 2:52 pm

(a) When the pH of 0.10 M HClO2(aq) was measured, it was found to be 1.2. What are the values of Ka and pKa of chlorous acid?

To find [H3O+], you take the antilog of the pH, and [H3O+] also equals [ClO2-]. For the Ka equation, Ka = [H3O+][ClO2-]/[HClO2]. How do you know that [HClO2] = (.10 - .06) when you don't know if the concentrations of [H3O+] and [ClO2-} are at equilibrium?

John Liang 2I
Posts: 102
Joined: Fri Aug 30, 2019 12:18 am

Re: 6D.3a

Postby John Liang 2I » Sun Jan 19, 2020 3:01 pm

JamieVu_2C wrote:(a) When the pH of 0.10 M HClO2(aq) was measured, it was found to be 1.2. What are the values of Ka and pKa of chlorous acid?

To find [H3O+], you take the antilog of the pH, and [H3O+] also equals [ClO2-]. For the Ka equation, Ka = [H3O+][ClO2-]/[HClO2]. How do you know that [HClO2] = (.10 - .06) when you don't know if the concentrations of [H3O+] and [ClO2-} are at equilibrium?


.06 is indeed the concentrations of both products at equilibrium. When the weak acid is in solution, it is already dissociating. Therefore, you can subtract .06 from .1 to find out how much of the initial HClO2 was converted into H3O+ and ClO2-.


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