Animated Solution for Chemistry - Ionic Equilibrium: Dilution process of different aqueous solutions; with water, are given in List-I. The effects of dilution of the solutions on [H+] are given in List-II.
(Note : Degree of dissociation (α) of weak acid and weak base is <<1; degree of hydrolysis of salt <<1; [H+] represents the concentration of H+ ions)
List-I
(P)
(10 mL of 0.1 M NaOH + 20 mL of 0.1 M acetic acid) diluted to 60 mL
(Q)
(20 mL of 0.1 M NaOH + 20 mL of 0.1 M acetic acid) diluted to 80 mL
(R)
(20 mL of 0.1 M HCl + 20 mL of 0.1 M ammonia solution) diluted to 80 mL
(S)
10 mL saturated solution of Ni(OH)2 in equilibrium with excess solid Ni(OH)2 is diluted to 20 mL (solid Ni(OH)2 is still present after dilution).
List-II
(1)
the value of [H+] does not change on dilution
(2)
the value of [H+] changes to half of its initial value on dilution
(3)
the value of [H+] changes to two times of its initial value on dilution
(4)
the value of [H+] changes to 21 times of its initial value on dilution
(5)
the value of [H+] changes to 2 times of its initial value on dilution
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Visualized Solution
Dilution of Aqueous Solutions
We need to evaluate the effect of dilution on the hydrogen ion concentration, [H+], for four different aqueous systems.
Case P: Acidic Buffer Formation
Initial: 10 mL of 0.1 M NaOH+20 mL of 0.1 M CH3COOH
nNaOH=10×0.1=1 mmol
nCH3COOH=20×0.1=2 mmol
After reaction: 1 mmol CH3COONa and 1 mmol CH3COOH remain.
Case P: Effect of Dilution
For an acidic buffer: [H+]=Ka[Salt][Acid]
[H+]=Kansalt/Vnacid/V=Kansaltnacid
Dilution changes V, but not the ratio of moles.
Therefore, [H+] does not change. (P → 1)
Case Q: Salt of Weak Acid and Strong Base
Initial: 20 mL of 0.1 M NaOH+20 mL of 0.1 M CH3COOH
nNaOH=2 mmol,nCH3COOH=2 mmol
Complete neutralization forms 2 mmol of CH3COONa (Salt of WA-SB).
Case Q: Hydrolysis Formula
For WA-SB salt, [OH−]=KaKwC
[H+]=[OH−]Kw=KaKwCKw=CKwKa
⇒[H+]∝C1
Case Q: Effect of Dilution
Initial volume V1=40 mL, Final volume V2=80 mL.
Volume is doubled ⇒ Concentration is halved (C2=C1/2).
[H+]1[H+]2=C2C1=C1/2C1=2
Therefore, [H+] changes to 2 times. (Q → 5)
Case R: Salt of Strong Acid and Weak Base
Initial: 20 mL of 0.1 M HCl+20 mL of 0.1 M NH3
nHCl=2 mmol,nNH3=2 mmol
Complete neutralization forms 2 mmol of NH4Cl (Salt of SA-WB).
Case R: Hydrolysis Formula
For SA-WB salt, [H+]=KbKwC
⇒[H+]∝C
Case R: Effect of Dilution
Initial volume V1=40 mL, Final volume V2=80 mL.
Volume is doubled ⇒ Concentration is halved (C2=C1/2).
[H+]1[H+]2=C1C2=C1C1/2=21
Therefore, [H+] changes to 21 times. (R → 4)
Case S: Saturated Solution with Excess Solid
System: Ni(OH)2(s)⇌Ni2+(aq)+2OH−(aq)
Ksp=[Ni2+][OH−]2
Dilution from 10 mL to 20 mL with solid still present.
Case S: Effect of Dilution
Since solid is present, the solution remains saturated.
Concentrations [Ni2+] and [OH−] remain constant to satisfy Ksp.
If [OH−] is constant, [H+]=[OH−]Kw is also constant.
Therefore, [H+] does not change. (S → 1)
Final Matrix Match
P → 1
Q → 5
R → 4
S → 1
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The Sigma Insight: Hydrolsis of Salts
Solution Diagram
The Beauty of Dilution
Imagine you have a beaker filled with a chemical solution. You add some pure water to it. What happens? Intuitively, you might think that everything just gets "weaker" or less concentrated. But in the fascinating world of ionic equilibrium, dilution is not a one-size-fits-all process!
Depending on what exactly is dissolved in that beaker—a buffer, a hydrolyzed salt, or a saturated solid—the hydrogen ion concentration, [H+], can react in completely different ways.
In this epic problem, we are going to explore four distinct cases. We will mix reagents, identify the resulting system, and then mathematically prove exactly how dilution affects the pH. Let's dive in!
Case P
The Resilient Buffer
In our first scenario, we mix 10 mL of 0.1 M NaOH with 20 mL of 0.1 M CH3COOH.
Before we even think about dilution, we must figure out what is actually in the beaker. We have 1 mmol of a strong base and 2 mmol of a weak acid. They will react! The strong base will completely consume 1 mmol of the weak acid to form 1 mmol of sodium acetate (CH3COONa).
What remains? We have 1 mmol of unreacted acetic acid and 1 mmol of the newly formed salt. This mixture of a weak acid and its conjugate base forms an acidic buffer.
Now, how does a buffer respond to dilution? The hydrogen ion concentration is governed by the equation:
[H+]=Ka[Salt][Acid]
Notice that both the acid and the salt are in the exact same volume of solution. If we rewrite concentration as moles divided by volume, the volume term completely cancels out!
[H+]=Kansaltnacid
When we add water to dilute the solution from 30 mL to 60 mL, the volume changes, but the ratio of the moles remains perfectly untouched. Therefore, the [H+] of a buffer does not change on dilution. Case P matches with 1.
Case Q
The Basic Salt
Next, we mix 20 mL of 0.1 M NaOH with 20 mL of 0.1 M CH3COOH.
This time, we have exactly 2 mmol of strong base and 2 mmol of weak acid. They perfectly neutralize each other! The only thing left in the beaker is 2 mmol of sodium acetate.
Sodium acetate is a salt of a weak acid and a strong base. It undergoes anionic hydrolysis, making the solution basic. The hydroxide ion concentration is given by:
[OH−]=KaKwC
Using the ionic product of water (Kw=[H+][OH−]), we can find the hydrogen ion concentration:
[H+]=CKwKa
Look closely at this relationship. The [H+] is inversely proportional to the square root of the concentration C.
When we dilute the solution from 40 mL to 80 mL, we are doubling the volume. Doubling the volume means the concentration C is halved (C2=C1/2).
Because of the inverse square root relationship, halving the concentration increases the [H+] by a factor of 2. Thus, Case Q matches with 5.
Case R
The Acidic Salt
In our third scenario, we mix 20 mL of 0.1 M HCl with 20 mL of 0.1 M NH3.
Here, we have 2 mmol of a strong acid and 2 mmol of a weak base. They completely neutralize to form 2 mmol of ammonium chloride (NH4Cl).
This is a salt of a strong acid and a weak base, which undergoes cationic hydrolysis, making the solution acidic. The formula for the hydrogen ion concentration is:
[H+]=KbKwC
This time, the [H+] is directly proportional to the square root of the concentration C.
Just like in Case Q, we are diluting the solution from 40 mL to 80 mL. The volume is doubled, so the concentration C is halved.
Because of the direct square root relationship, halving the concentration decreases the [H+] by a factor of 2. It becomes 21 times its initial value. Therefore, Case R matches with 4.
Case S
The Saturated Equilibrium
Finally, we have a saturated solution of nickel hydroxide, Ni(OH)2, in equilibrium with its solid form.
The problem gives us a massive hint: even after diluting the solution from 10 mL to 20 mL, solid Ni(OH)2 is still present.
Why is this so important? Because as long as the solid is present, the solution remains perfectly saturated. The equilibrium is governed by the solubility product:
Ksp=[Ni2+][OH−]2
When we add water, the system simply dissolves a little bit more of the solid to maintain the exact same equilibrium concentrations. The [OH−] remains absolutely constant.
And if the hydroxide concentration is constant, the hydrogen ion concentration must also be constant! Therefore, the [H+] does not change on dilution. Case S matches with 1.
The Final Verdict
By carefully analyzing the chemical nature of each mixture, we successfully decoded the effects of dilution. Buffers resist change, hydrolyzed salts shift their pH based on square root relationships, and saturated solutions hold their ground as long as solid remains.
