The solubility of a salt of weak acid(AB) at pH 3 is Y x 10^-3 mol L^-1. The value of Y is_______.
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Q: Why does the solubility increase in an acidic medium?

According to Le Chatelier's principle, the $H^+$ ions react with $B^-$ to form $HB$. This removes $B^-$ from the product side of the dissolution equilibrium, driving the reaction forward and dissolving more salt.

Q: Why is $[A^+]$ equal to $S$ but $[B^-]$ is not?

$A^+$ is a spectator ion in the acid-base reaction; it doesn't react with anything, so its concentration represents the total amount of salt dissolved. $B^-$, however, reacts with $H^+$, so its total dissolved amount is split between free $B^-$ and protonated $HB$.

Q: Can we always ignore the '1' in the bracket $(1 + [H^+]/K_a)$?

No! You can only ignore it when $[H^+]/K_a \gg 1$, which happens when the solution is highly acidic compared to the $pK_a$ of the weak acid. Always check the values before approximating.

Q: What if the salt was of a strong acid, like NaCl?

For a salt of a strong acid, the anion (like $Cl^-$) does not undergo hydrolysis. The term $[H^+]/K_a$ would be practically zero, and the solubility would simply be $\sqrt{K_{sp}}$, unaffected by pH.

Q: How did we calculate $\sqrt{2 \times 10^{-5}}$ without a calculator?

The trick is to make the power of 10 even so it can be easily halved out of the square root. We rewrote $2 \times 10^{-5}$ as $20 \times 10^{-6}$. Then, $\sqrt{10^{-6}} = 10^{-3}$, and $\sqrt{20}$ is between $\sqrt{16}$ (4) and $\sqrt{25}$ (5), roughly 4.47.

Question

The solubility of a salt of weak acid(AB) at pH 3 is $Y 	imes 10^{-3} 	ext{ mol L}^{-1}$. The value of Y is_______.
(Given that the value of solubility product of AB ($K_{sp}$) = $2 	imes 10^{-10}$ and the value of ionization constant of HB($K_a$) = $1 	imes 10^{-8}$)

Sigma Percentile · Accepted Answer

### The Illusion of Simple Solubility

When we think of solubility, we usually just write down the $K_{sp}$ equation and call it a day. But what happens when the salt is born from a weak acid? The rules of the game change completely.

Imagine you are standing on the edge of a beaker containing the sparingly soluble salt $AB$. In pure water, it establishes a simple equilibrium:

$$AB(s) ightleftharpoons A^+(aq) + B^-(aq)$$

But this solution is not pure water; it has a $pH$ of $3$, meaning it is swimming with $H^+$ ions.

### The Dance of Simultaneous Equilibria

Here is where the magic happens. The anion $B^-$ is the conjugate base of a weak acid $HB$. It has a strong affinity for protons. As soon as $B^-$ enters the solution, the abundant $H^+$ ions attack it, forming undissociated $HB$:

$$B^-(aq) + H^+(aq) ightleftharpoons HB(aq)$$

**According to Le Chatelier's principle**, this secondary reaction continuously consumes $B^-$ ions. To compensate for this loss, the primary dissolution equilibrium shifts forward. The salt dissolves more than it normally would! This is the essence of simultaneous equilibria.

### Deriving the Master Equation

To solve this, we need to track where all the dissolved salt goes. If the total solubility is $S$, then the concentration of the spectator ion $A^+$ is exactly $S$.

However, the $B^-$ ions are split into two forms: free $B^-$ and protonated $HB$. This gives us our mass balance equation:

$$S = [B^-] + [HB]$$

We know from the acid dissociation constant $K_a$ that:

$$K_a = \frac{[H^+][B^-]}{[HB]} \implies [HB] = \frac{[H^+][B^-]}{K_a}$$

Substituting this into our mass balance gives:

$$S = [B^-] + \frac{[H^+][B^-]}{K_a} = [B^-] \left( 1 + \frac{[H^+]}{K_a} ight)$$

Now, we isolate $[B^-]$:

$$[B^-] = \frac{S}{1 + \frac{[H^+]}{K_a}}$$

Finally, we plug this back into the $K_{sp}$ expression:

$$K_{sp} = [A^+][B^-] = S \cdot \frac{S}{1 + \frac{[H^+]}{K_a}}$$

Rearranging this yields our beautiful master equation for solubility in an acidic medium:

$$S = \sqrt{K_{sp} \left( 1 + \frac{[H^+]}{K_a} ight)}$$

### Crunching the Numbers

Now, let's bring back the values we found earlier. We have $K_{sp} = 2 	imes 10^{-10}$, $K_a = 10^{-8}$, and $[H^+] = 10^{-3} 	ext{ M}$.

Substituting these into our master equation:

$$S = \sqrt{2 	imes 10^{-10} \left( 1 + \frac{10^{-3}}{10^{-8}} ight)}$$

Look at the term inside the parenthesis: $\frac{10^{-3}}{10^{-8}} = 10^5$. 
Since $10^5$ is $100,000$, adding $1$ to it makes practically no difference. **We can safely approximate $1 + 10^5 \approx 10^5$.**

$$S \approx \sqrt{2 	imes 10^{-10} 	imes 10^5} = \sqrt{2 	imes 10^{-5}}$$

To make the square root easy to calculate without a calculator, we adjust the power of $10$ to an even number:

$$S = \sqrt{20 	imes 10^{-6}} = \sqrt{20} 	imes 10^{-3}$$

Since $\sqrt{16} = 4$ and $\sqrt{25} = 5$, $\sqrt{20}$ is right in the middle, approximately $4.47$.

$$S = 4.47 	imes 10^{-3} 	ext{ M}$$

Comparing this to the given format $Y 	imes 10^{-3} 	ext{ M}$, we find our final answer:

**$Y = 4.47$**

This problem is a masterpiece of physical chemistry, beautifully weaving together solubility products, acid-base equilibria, and smart mathematical approximations!

Enter Numerical Value:

Visualized Solution

$p H = 3 ⟹ [H^{+}] = 1 0^{- 3} M$

$K_{s p} = [A^{+}] [B^{-}]$

$K_{a} = \frac{[ H ^{+} ] [ B ^{-} ]}{[ H B ]}$

$S = [B^{-}] + [H B]$

$[B^{-}] = \frac{S}{1 + \frac{[ H ^{+} ]}{K _{a}}}$

$S^{2} = K_{s p} (1 + \frac{[ H ^{+} ]}{K _{a}})$

$S = 2 \times 1 0^{- 10} (1 + \frac{1 0 ^{- 3}}{1 0 ^{- 8}})$

$1 + 1 0^{5} \approx 1 0^{5}$

$S = 20 \times 1 0^{- 6}$

$Y = 4.47$

The Sigma Insight: Solubility Product and Common Ion Effect

The Illusion of Simple Solubility

The Dance of Simultaneous Equilibria

Deriving the Master Equation

Crunching the Numbers

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Enter Numerical Value:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } pH=3⟹[H+]=10−3 M

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Ksp​=[A+][B−]

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Ka​=[HB][H+][B−]​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } S=[B−]+[HB]

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } [B−]=1+Ka​[H+]​S​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } S2=Ksp​(1+Ka​[H+]​)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } S=2×10−10(1+10−810−3​)​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 1+105≈105

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } S=20×10−6​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Y=4.47