6D13: calculating pH

[annabelchen2a]

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]

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]

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

[Hannah_Butler_2E]

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]

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.