Question

Calculate \(\left[\mathrm{OH}^{-}\right]\)and \(\mathrm{pH}\) for each of the following strong base solutions: (a) \(0.182 \mathrm{M} \mathrm{KOH}\), (b) \(3.165 \mathrm{~g}\) of KOH in \(500.0 \mathrm{~mL}\) of solution, (c) \(10.0 \mathrm{~mL}\) of \(0.0105 \mathrm{M} \mathrm{Ca}(\mathrm{OH})_{2}\) diluted to \(500.0 \mathrm{~mL}\), (d) a solution formed by mixing \(20.0 \mathrm{~mL}\) of \(0.015 \mathrm{M} \mathrm{Ba}(\mathrm{OH})_{2}\) with \(40.0 \mathrm{~mL}\) of \(8.2 \times 10^{-3} \mathrm{M} \mathrm{NaOH}\).

Short answer

(a) For 0.182 M KOH: \[\left[\mathrm{OH}^{-}\right] = 0.182 \mathrm{M}\], \(\mathrm{pOH} = -\log{(0.182)}\), and \(\mathrm{pH} = 14 - \mathrm{pOH}\). (b) For 3.165 g of KOH in 500.0 mL of solution: \[\left[\mathrm{OH}^{-}\right] = 0.1128\, \text{M}\], \(\mathrm{pOH} = -\log{(0.1128)}\), and \(\mathrm{pH} = 14 - \mathrm{pOH}\). (c) For 10.0 mL of 0.0105 M Ca(OH)₂ diluted to 500.0 mL: \[\left[\mathrm{OH}^{-}\right] = 0.00042 \, \text{M}\], \(\mathrm{pOH} = -\log{(0.00042)}\), and \(\mathrm{pH} = 14 - \mathrm{pOH}\). (d) A solution formed by mixing 20.0 mL of 0.015 M Ba(OH)₂ with 40.0 mL of 8.2 x 10⁻³ M NaOH: \[\left[\mathrm{OH}^{-}\right] = 0.015467 \, \text{M}\], \(\mathrm{pOH} = -\log{(0.015467)}\), and \(\mathrm{pH} = 14 - \mathrm{pOH}\).

Step-by-step solution

Calculate OH⁻ concentration

Since KOH is a strong base, the concentration of hydroxide ions is equal to the concentration of KOH: \(\left[\mathrm{OH}^{-}\right] = 0.182 \mathrm{M}\)

Calculate pOH

Now, we will calculate the pOH: \(\mathrm{pOH} = -\log{(0.182)}\)

Calculate pH

Finally, we will calculate the pH of the solution: \(\mathrm{pH} = 14 - \mathrm{pOH}\) (b) 3.165 g of KOH in 500.0 mL of solution

Calculate KOH concentration

First, we need to calculate the concentration of KOH in the solution. The molecular weight of KOH is 39.10 g/mol (K) + 15.999 g/mol (O) + 1.008 g/mol (H) = 56.11 g/mol. The number of moles of KOH is: $$\frac{3.165\,\text{g}}{56.11\,\text{g/mol}}=0.0564 \,\text{mol}$$ The concentration of KOH is: $$\frac{0.0564\,\text{mol}}{0.500\,\text{L}}=0.1128\, \text{M}$$

Calculate OH⁻ concentration

Since KOH is a strong base, the concentration of hydroxide ions is equal to the concentration of KOH: \(\left[\mathrm{OH}^{-}\right] = 0.1128\, \text{M}\)

Calculate pOH

Now, we will calculate the pOH: \(\mathrm{pOH} = -\log{(0.1128)}\)

Calculate pH

Finally, we will calculate the pH of the solution: \(\mathrm{pH} = 14 - \mathrm{pOH}\) (c) 10.0 mL of 0.0105 M Ca(OH)₂ diluted to 500.0 mL

Calculate diluted Ca(OH)₂ concentration

First, we need to calculate the concentration of Ca(OH)₂ after dilution using the formula: $$C1V1 = C2V2$$ Where C1 is the initial concentration, V1 is the initial volume, C2 is the final concentration, and V2 is the final volume. $$0.0105\,\text{M} \times 0.010\,\text{L} = C2 \times 0.500\,\text{L}$$ Solving for C2, we get: $$C2 = 0.00021\,\text{M}$$

Calculate OH⁻ concentration

Since Ca(OH)₂ produces 2 moles of hydroxide ions per mole of Ca(OH)₂, the concentration of hydroxide ions in the diluted solution is: $$\left[\mathrm{OH}^{-}\right] = 2 \times 0.00021\,\text{M} = 0.00042\,\text{M}$$

Calculate vOH

Next, we will calculate the pOH: \(\mathrm{pOH} = -\log{(0.00042)}\)

Calculate pH

Finally, we will calculate the pH of the solution: \(\mathrm{pH} = 14 - \mathrm{pOH}\) (d) Mixing 20.0 mL of 0.015 M Ba(OH)₂ with 40.0 mL of 8.2 x 10⁻³ M NaOH

Calculate total OH⁻ moles

First, we need to find the total moles of hydroxide ions from the solutions: For Ba(OH)₂, moles of OH⁻ = 2 (since Ba(OH)₂ releases 2 OH⁻ ions per formula unit) x 0.015 M x 0.020 L = 0.0006 mol For NaOH, moles of OH⁻ = 8.2 x 10⁻³ M x 0.040 L = 0.000328 mol Total moles of OH⁻ = 0.0006 mol + 0.000328 mol = 0.000928 mol

Calculate OH⁻ concentration

Now we need to find the concentration of hydroxide ions in the mixture: $$\left[\mathrm{OH}^{-}\right] = \frac{0.000928\,\text{mol}}{0.020\,\text{L} + 0.040\,\text{L}} = 0.000928\,\text{mol}/0.060\,\text{L} = 0.015467\,\text{M}$$

Calculate pOH

Now, we will calculate the pOH: \(\mathrm{pOH} = -\log{(0.015467)}\)

Calculate pH

Finally, we will calculate the pH of the solution: \(\mathrm{pH} = 14 - \mathrm{pOH}\)

Key concepts

Strong Bases

Strong bases are compounds that completely dissociate in water to produce hydroxide ions (\( \text{OH}^- \)). This property allows them to significantly increase the pH of a solution.
The more a base dissociates, the stronger it is considered. Some common examples of strong bases include potassium hydroxide (KOH), sodium hydroxide (NaOH), and barium hydroxide (\( \text{Ba(OH)}_2 \)).
These compounds dissolve fully, meaning that if you put KOH in water, all of it turns into \( \text{K}^+ \) ions and hydroxide ions (\( \text{OH}^- \)).

• The complete dissociation characteristic of strong bases gives them their potent ability to raise the pH.
• As seen in \( \text{Ca(OH)}_2 \), which releases two hydroxide ions per formula unit, the more \( \text{OH}^- \) produced, the more the change in pH.
• Understanding the behavior of strong bases is essential for pH calculations and determining solution concentration.

Hydroxide Ion Concentration

To determine the hydroxide ion concentration in a solution containing a strong base, you can often start with the concentration of the base itself.
For instance, in a solution of KOH, the concentration of \( \text{OH}^- \) ions is equal to the concentration of KOH if it is completely dissolved in water.
However, in bases like \( \text{Ca(OH)}_2 \), which release two \( \text{OH}^- \) ions per molecule, the concentration of \( \text{OH}^- \) is double that of the base.
When performing calculations:

• Using the formula \( n = M \times V \) can help find the moles of hydroxide ions, where \( n \) is the number of moles, \( M \) is molarity, and \( V \) is volume in liters.
• For solutions mixed or diluted, sum the moles of hydroxide ions to find the total concentration in the final volume.
• Knowing the hydroxide concentration helps calculate both \( \text{pOH} \) and subsequently \( \text{pH} \)

Dilution of Solutions

Dilution involves reducing the concentration of a solute in a solution, typically by adding more solvent.
When you dilute a strong base, you maintain the total number of moles of \( \text{OH}^- \), but the concentration decreases due to the increased volume.
The formula used is: \[ C_1 \cdot V_1 = C_2 \cdot V_2 \] Where \( C_1 \) and \( V_1 \) are the initial concentration and volume, while \( C_2 \) and \( V_2 \) are the final concentration and volume.

• This equation ensures that the total amount of solute remains the same before and after dilution.
• When considering dilutions in strong bases, it is vital to remember that the dilution primarily affects concentration but not the total amount of substance.
• After finding the new concentration, the hydroxide concentration helps further calculate the pH by finding \( \text{pOH} \).