Consider a sturated solution of silver chloride that is in contact with solid silver chloride. The solubility equilibrium can be represented as
`AgCl(s)hArrAg^(+)(aq.)+Cl^(-)(aq.)," "K_(sp)=[Ag^(+)(aq.)][Cl^(-)(aq.)]`
Where `K_(sp)` is clled the solubility product constant or simply the solubility product. In general, the solubility product of a compound is the product of the molar concentrations of the constituent ions, each raised to the power of its stoichiometric coefficient in the equilibrium equation.
For concentrations of ions that do not necessarliy correpond to equilibrium conditions we use the reaction quotient (Q) which is clled the ion or ionic prodect (Q) to predict whether a precipitate will from. Note that (Q) has the same for as `K_(sp)` are
`QltK_(sp)` Unsaturated solution
`Q=K_(sp)` Saturated solution
`Qgt_(sp)` Supersaturated solution, precipitate will from
At `25^(@)C,` will a precipitate of Mg `(OH)_(2)` from when a 0.0001 M solution of `Mg(NO_(3))_(2)` is adjusted to a pH of 9.0 ? At what minimum value of pH will precipition start ?
`["Given" : K_(sp)(Mg(OH)_(2))=10^(-11)M^(3)]`

Options

A. No,pH=3.5
B. No pH 10.5
C. No,pH=6.0
D. Yes, pH=8.5

Answer

To solve the problem, we need to determine whether a precipitate of magnesium hydroxide (Mg(OH)₂) will form when a 0.0001 M solution of magnesium nitrate (Mg(NO₃)₂) is adjusted to a pH of 9.0. We will also find the minimum pH at which precipitation begins.

### Step 1: Determine the concentration of hydroxide ions (OH⁻) at pH 9.0
The pH is given as 9.0. We can calculate the pOH using the formula:
\[ \text{pOH} = 14 - \text{pH} \]
Substituting the value:
\[ \text{pOH} = 14 - 9 = 5 \]
Now, we can find the concentration of hydroxide ions (OH⁻) using the formula:
\[ \text{[OH⁻]} = 10^{-\text{pOH}} = 10^{-5} \, \text{M} \]

### Step 2: Calculate the concentration of magnesium ions (Mg²⁺)
The magnesium nitrate dissociates in solution as follows:
\[ \text{Mg(NO₃)₂} \rightarrow \text{Mg}^{2+} + 2\text{NO₃}^{-} \]
Given that the concentration of Mg(NO₃)₂ is 0.0001 M, the concentration of Mg²⁺ ions will also be:
\[ \text{[Mg}^{2+}\text{]} = 0.0001 \, \text{M} = 10^{-4} \, \text{M} \]

### Step 3: Write the expression for the solubility product (Ksp)
The solubility product constant (Ksp) for magnesium hydroxide is given as:
\[ K_{sp}(\text{Mg(OH)}_2) = [\text{Mg}^{2+}][\text{OH}^-]^2 \]
Substituting the concentrations we have:
\[ K_{sp} = (10^{-4})(10^{-5})^2 = (10^{-4})(10^{-10}) = 10^{-14} \]

### Step 4: Compare Q with Ksp to determine precipitation
Now, we need to compare the calculated Q (reaction quotient) with Ksp:
\[ Q = [\text{Mg}^{2+}][\text{OH}^-]^2 = (10^{-4})(10^{-5})^2 = 10^{-14} \]
Since \( Q = K_{sp} \), the solution is saturated, and no precipitate will form at pH 9.0.

### Step 5: Find the minimum pH for precipitation to start
To find the minimum pH at which precipitation starts, we need to set Q equal to Ksp and solve for the concentration of OH⁻ that would cause precipitation:
\[ K_{sp} = [\text{Mg}^{2+}][\text{OH}^-]^2 \]
Let [OH⁻] = x, then:
\[ 10^{-11} = (10^{-4})(x^2) \]
Solving for x:
\[ x^2 = \frac{10^{-11}}{10^{-4}} = 10^{-7} \]
\[ x = 10^{-3.5} = 3.16 \times 10^{-4} \, \text{M} \]
Now, we can calculate the pOH:
\[ \text{pOH} = -\log(3.16 \times 10^{-4}) \approx 3.5 \]
Finally, we find the corresponding pH:
\[ \text{pH} = 14 - \text{pOH} = 14 - 3.5 = 10.5 \]

### Conclusion
1. At pH 9.0, no precipitate of Mg(OH)₂ will form.
2. The minimum pH at which precipitation will start is **10.5**.