# I understand you compare Ka values to determine the pH but this order just seems to be all over the place? HCl: strong acid so low pH C6H5NH3Cl: Ka=1.35x10^-5 NH4+ Ka=5.56x10^-10 MgCl2=neutral HClO2Ka=1.2x10^-2 HOCl Ka=4.0x10^-8

Apr 8, 2018

Some of these are tricky because the problem gives you the acid ${K}_{A}$, but they are actually asking for a 0.25M solution of the conjugate base. (NaOCl and $N a C l {O}_{2}^{-}$)

#### Explanation:

In the case of the two species that have sodium (above), you have to use the ${K}_{B}$ value, and this is found by ${K}_{B} = {K}_{W} / {K}_{A}$, so that the weaker the acid is, the stronger its conjugate base will be.

HCl comes apart, so that will be lowest pH.
The next options on the list for lowest pH are $N H {4}^{+}$ and C6H5NH3Cl. C6H5NH3Cl has a smaller ${K}_{A}$, so it is next up. Then $N {H}_{4}^{+}$.

$M g C {l}_{2}$, NaOCl and $N a C l {O}_{2}^{-}$ are the ones you have left. $M g C {l}_{2}$ is salty water, so its pH will be about 7.

NaOCl and $N a C l {O}_{2}^{-}$ are the last two. $H C l {O}_{2}$ is a weak acid, but it is a stronger acid than HClO, so it will have a weaker conjugate base.

HOCl is a weaker acid, so it will have a stronger conjugate base.

This means NaOCl will make a more basic solution than will $N a C l {O}_{2}^{-}$

I hope that helps.