Which Salt Produces Lowest pH Solution?

[Soaring Crane]

Which one of the following salts, when dissolved in water, produces the solution with the lowest pH?


a.NaCl

b.NH4Cl

c.MgCl2

d.AlCl3

Ammonium chloride

My answer lies with b or d since they are both acidic salts, but I think is it d since the K_a is greater for Al(3+) than NH4+.


Thanks.

 

[siddharth]

The general rule of thumb is that the salt of a

• Strong Acid - Strong Base is neutral
• Weak Acid - Strong Base is basic
• Strong Acid - Weak Base is acidic
• Weak acid - Weak base depends on ka and kb

 

[Soaring Crane]

So the base for NH4Cl is NH4(OH).
The base for AlCl3 is Al(OH)3.

Both spawn from the strong acid HCl, so how do I rank them from here with this criteria?

 

[siddharth]

So the base for NH4Cl is NH4(OH).
The base for AlCl3 is Al(OH)3.

Right

Both spawn from the strong acid HCl, so how do I rank them from here with this criteria?

In this case, look at the equilibrium which exists.

R⁺ + H₂O <====> ROH + H⁺
where R⁺ can be NH₄⁺ or Al³⁺.

The challenge here is to relate the pH to Kb. You can do this mathematically and get a relationship from which you can get the answer. If you assume the initial concentration of R⁺ to be 'a' and the extent of dissociation to be 'x', can you come up with a equation relating the pH and Kb?
(Hint: Try assuming that x is negligible when compared to 1 to simplify your calculations)

 

[Soaring Crane]

pH = 14 + log (K_b*[A-]/[HA]), where K_b = [HA][OH-]/[A-]??

Is this an expression for that relationship?

 

[siddharth]

pH = 14 + log (K_b*[A-]/[HA]), where K_b = [HA][OH-]/[A-]??

Is this an expression for that relationship?

Not quite.
First of all can you see that $$k_h = \frac{k_w}{k_b}$$? This is because

H₂O <=======> H⁺ + OH^- ---- k_w

ROH <=====> R⁺ + OH^- ----- k_b

Now if you subtract the two equations above you get,
R⁺ + H₂O <====> ROH + H⁺
which is your hydrolysis equilibrium. So K for the above equilibrium will be $$k_h = \frac{k_w}{k_b}$$.

If the initial concentration of R+ is 'a' and the extent of dissociation is 'x', then at equilibrium, the concentration of ROH and H⁺ will be ax and the concentration of R⁺ will be a-ax. Is it clear till this?

Now, you know that

$$k_h = \frac{k_w}{k_b} = \frac{(ROH)(H^+)}{(R^+)}$$.

So substitute the concentrations in the above equation. Remember, you need the concentration of H⁺ (ie, 'ax'). Can you take it from here? If any of the above isn't clear, don't hesitate to ask.