6D13: calculating pH

Moderators: Chem_Mod, Chem_Admin

annabelchen2a
Posts: 101
Joined: Wed Sep 30, 2020 10:01 pm

6D13: calculating pH

Postby annabelchen2a » Mon Jan 18, 2021 11:37 am

Rank the following solutions in order of increasing pH: (a) 1.0 * 10^-5M HCl(aq); (b) 0.20M CH3NH3Cl(aq); (c) 0.20M CH3COOH(aq); (d) 0.20M C6H5NH2(aq). Justify your ranking.

I got the right answer by using the Ka and Kb values (from tables 6C.1 and 6C.2) for b, c, and d to calculate their respective pH values and just calculating pH = -log(1*10^-5) for (a).

Is this the way we're supposed solve this problem? Is there any way to solve this problem without Ka or Kb values?

Selena Quispe 2I
Posts: 124
Joined: Wed Sep 30, 2020 10:01 pm
Been upvoted: 3 times

Re: 6D13: calculating pH

Postby Selena Quispe 2I » Mon Jan 18, 2021 12:20 pm

Yes! You answered this question correctly! I think we need the ka values to solve for pH as when you use the ka we are finding the concentration of hydronium which is needed when calculating the pH (-log[H3O+]). I hope this helps!

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

Re: 6D13: calculating pH

Postby Joseph Hsing 2C » Mon Jan 18, 2021 12:21 pm

I think you would need Ka and Kb values since most of the given compounds are weak acids and bases.

Hannah_Butler_2E
Posts: 41
Joined: Sat Jul 20, 2019 12:16 am

Re: 6D13: calculating pH

Postby Hannah_Butler_2E » Mon Jan 18, 2021 4:06 pm

annabelchen2a wrote:Rank the following solutions in order of increasing pH: (a) 1.0 * 10^-5M HCl(aq); (b) 0.20M CH3NH3Cl(aq); (c) 0.20M CH3COOH(aq); (d) 0.20M C6H5NH2(aq). Justify your ranking.

I got the right answer by using the Ka and Kb values (from tables 6C.1 and 6C.2) for b, c, and d to calculate their respective pH values and just calculating pH = -log(1*10^-5) for (a).

Is this the way we're supposed solve this problem? Is there any way to solve this problem without Ka or Kb values?


How did you find the Ka value for (b) CH3NH3Cl ? I can't find it in the table and when I tried just solving assuming it was a strong acid I got the wrong answer.

annabelchen2a
Posts: 101
Joined: Wed Sep 30, 2020 10:01 pm

Re: 6D13: calculating pH

Postby annabelchen2a » Mon Jan 18, 2021 6:21 pm

Hannah_Butler_2E wrote:How did you find the Ka value for (b) CH3NH3Cl ? I can't find it in the table and when I tried just solving assuming it was a strong acid I got the wrong answer.

I found the Kb value for CH3NH2 in table 6C.2, and got a Ka value of 2.778 * 10^-11 from that.


Return to “Equilibrium Constants & Calculating Concentrations”

Who is online

Users browsing this forum: No registered users and 1 guest