Question

Calculate \(\left[\mathrm{OH}^{-}\right]\)and \(\mathrm{pH}\) for (a) \(1.5 \times 10^{-3} \mathrm{M} \mathrm{Sr}(\mathrm{OH})_{2}\), (b) \(2.250 \mathrm{~g}\) of \(\mathrm{LiOH}\) in \(250.0 \mathrm{~mL}\) of solution, (c) \(1.00 \mathrm{~mL}\) of \(0.175 \mathrm{MNaOH}\) diluted to \(2.00 \mathrm{~L}\) (d) a solution formed by adding \(5.00 \mathrm{~mL}\) of \(0.105 \mathrm{M} \mathrm{KOH}\) to \(15.0 \mathrm{~mL}\) of \(9.5 \times 10^{-2} \mathrm{M} \mathrm{Ca}(\mathrm{OH})_{2}\).

Short answer

The short answers for the hydroxide ion concentrations \(\left[\mathrm{OH}^{-}\right]\) and pH values are: (a) \(\left[\mathrm{OH}^{-}\right] = 3.0 \times 10^{-3} \mathrm{M}\) and pH = 11.478. (b) \(\left[\mathrm{OH}^{-}\right] = 0.200\,\text{M}\) and pH = 13.301. (c) \(\left[\mathrm{OH}^{-}\right] = 8.75 \times 10^{-5}\,\text{M}\) and pH = 9.942. (d) \(\left[\mathrm{OH}^{-}\right] = 0.170\,\text{M}\) and pH = 13.230.

Step-by-step solution

(a) Calculate the hydroxide ion concentration and pH of \(1.5 \times 10^{-3} \mathrm{M} \mathrm{Sr}(\mathrm{OH})_{2}\)

: Since Sr(OH)₂ is a strong base, it will completely dissociate in water, forming 2 moles of hydroxide ions for every mole of Sr(OH)₂: $$\mathrm{Sr}(\mathrm{OH})_{2} \rightarrow \mathrm{Sr}^{2+} + 2\mathrm{OH}^{-}.$$ So, multiply the molarity of Sr(OH)₂ by 2 to get the hydroxide ion concentration: $$\left[\mathrm{OH}^{-}\right] = 2 \times 1.5 \times 10^{-3} \mathrm{M} = 3.0 \times 10^{-3} \mathrm{M}.$$ Now, find the pOH and then the pH: $$\mathrm{pOH} = -\log\left(\left[\mathrm{OH}^{-}\right]\right) = -\log(3.0 \times 10^{-3}) \approx 2.522.$$ Since the pH of a solution can be determined by subtracting the pOH from 14: $$\mathrm{pH} = 14 - \mathrm{pOH} = 14 - 2.522 \approx 11.478.$$

(b) Calculate the hydroxide ion concentration and pH of \(2.250 \mathrm{~g}\) of \(\mathrm{LiOH}\) in \(250.0 \mathrm{~mL}\) of solution

: First, convert the mass of LiOH to moles using its molar mass (6.94 g/mol for Li and 15.999 g/mol for O and 1.008 g/mol for H): $$\frac{2.250\,\text{g}}{(6.94+15.999+1.008)\,\text{g/mol}} \approx 0.050\,\text{mol}\, \text{LiOH}.$$ Now, find the molarity of the solution: $$\frac{0.050\,\text{mol}}{0.250\,\text{L}} = 0.200\,\text{M}.$$ Since LiOH is a strong base, it dissociates completely, yielding one hydroxide ion per molecule: $$\mathrm{LiOH} \rightarrow \mathrm{Li}^{+} + \mathrm{OH}^{-}.$$ So, \(\left[\mathrm{OH}^{-}\right] = 0.200\,\text{M}\). Now, find the pOH and pH: $$\mathrm{pOH} = -\log\left(\left[\mathrm{OH}^{-}\right]\right) = -\log(0.200) \approx 0.699.$$ $$\mathrm{pH} = 14 - \mathrm{pOH} = 14 - 0.699 \approx 13.301.$$

(c) Calculate the hydroxide ion concentration and pH for a solution formed by diluting \(1.00 \mathrm{~mL}\) of \(0.175 \mathrm{M} \mathrm{NaOH}\) to \(2.00 \mathrm{~L}\)

: Calculate the moles of NaOH in the initial volume: $$0.175\,\text{M} \times 0.001\,\text{L} = 1.75 \times 10^{-4}\;\text{mol\,NaOH}.$$ Now, find the new molarity after dilution: $$\frac{1.75 \times 10^{-4}\,\text{mol}}{2.00\,\text{L}} = 8.75 \times 10^{-5}\,\text{M}.$$ Since NaOH is a strong base, it dissociates completely, yielding one hydroxide ion per molecule: $$\mathrm{NaOH} \rightarrow \mathrm{Na}^{+} + \mathrm{OH}^{-}.$$ So, \(\left[\mathrm{OH}^{-}\right] = 8.75 \times 10^{-5}\,\text{M}\). Now, find the pOH and pH: $$\mathrm{pOH} = -\log\left(\left[\mathrm{OH}^{-}\right]\right) = -\log(8.75 \times 10^{-5}) \approx 4.058.$$ $$\mathrm{pH} = 14 - \mathrm{pOH} = 14 - 4.058 \approx 9.942.$$

(d) Calculate the hydroxide ion concentration and pH of a solution formed by mixing \(5.00 \mathrm{~mL}\) of \(0.105 \mathrm{M} \mathrm{KOH}\) with \(15.0 \mathrm{~mL}\) of \(9.5 \times 10^{-2} \mathrm{M} \mathrm{Ca}(\mathrm{OH})_{2}\)

: First, find the moles of each hydroxide ion source: $$0.105\,\text{M} \times 0.005\,\text{L} = 5.25 \times 10^{-4}\;\text{mol\,KOH} \rightarrow 5.25 \times 10^{-4}\;\text{mol}\,\mathrm{OH}^{-}$$ $$9.5 \times 10^{-2}\,\text{M} \times 0.015\,\text{L} = 1.425 \times 10^{-3}\;\text{mol\,Ca(OH)_2} \rightarrow 2 \times 1.425 \times 10^{-3}\;\text{mol}\,\mathrm{OH}^{-} = 2.850 \times 10^{-3}\;\text{mol}\,\mathrm{OH}^{-}$$ Now, combine the moles of hydroxide ions and the total volume of the solution: $$\frac{5.25 \times 10^{-4} + 2.850 \times 10^{-3}}{0.005\,\text{L} + 0.015\,\text{L}} \approx 0.170\,\text{M} \;\mathrm{OH}^{-}.$$ Now, find the pOH and pH: $$\mathrm{pOH} = -\log\left(\left[\mathrm{OH}^{-}\right]\right) = -\log(0.170) \approx 0.770.$$ $$\mathrm{pH} = 14 - \mathrm{pOH} = 14 - 0.770 \approx 13.230.$$

Key concepts

Hydroxide Ion Concentration

The hydroxide ion concentration, denoted as \([\text{OH}^{-}]\), is a key factor when determining the basicity of a solution. The concentration of these ions provides direct insight into how basic or alkaline a solution is. Strong bases, such as Sr(OH)₂ and NaOH, completely dissociate in water. This dissociation releases hydroxide ions into the solution, fundamentally altering its pH level.
To calculate the hydroxide ion concentration, one starts by determining the number of hydroxide ions generated per mole of a base molecule. For instance, Sr(OH)₂ dissociates to produce two hydroxide ions per molecule, doubling the initial molarity of the base. Similarly, NaOH or LiOH in solution produces one hydroxide ion per molecule.
This directly influenced the exercises where we calculated \([\text{OH}^{-}]\) using the simple relationship: \[\text{OH}^{-} = n \times \text{Molarity of Base}\], where \(n\) represents the number of hydroxide ions released per molecule of base.

Strong Bases

Strong bases are compounds that completely dissociate into their ions in aqueous solutions. This complete dissociation is why they have such a significant impact on pH. Examples of strong bases include lithium hydroxide (LiOH), sodium hydroxide (NaOH), and calcium hydroxide (Ca(OH)₂).
These bases produce hydroxide ions, which significantly increase the \([\text{OH}^{-}]\) concentration of a solution, demonstrating their strength. Notably, the strong dissociation property makes it simpler to determine the hydroxide ion concentration since each molecule of the base contributes a consistent number of hydroxide ions.

• For instance, the dissolution of LiOH involves a straightforward conversion where each molecule yields one hydroxide ion.
• On the other hand, the reaction of Ca(OH)₂ with water releases two ions per molecule, exemplifying how different bases contribute varying amounts of OH⁻.

The exercises capitalizing on this concept showed the importance of translating the base's concentration into a count of hydroxide ions, simplifying pH calculations.

Chemical Equilibrium

In the context of strong bases, chemical equilibrium refers to the balance that occurs when a base dissolves in water, reaching near instantaneously as the base completely dissociates. Unlike weak bases, strong bases reach equilibrium without leaving undissociated base molecules. This allows for straightforward calculations of properties like pH based entirely on initial concentrations.
The dissociation equation, for instance, \[\text{NaOH} \rightarrow \text{Na}^+ + \text{OH}^-\], demonstrates that at equilibrium, all NaOH has become ions, with no reactant molecules left. In this equilibrium, only the products (ions) are significant.
This characteristic makes dealing with strong bases simpler in mathematical models and real-world applications, where predicting pH changes becomes far less complex as reflected in the provided exercise calculations.

Acid-Base Reaction

An acid-base reaction involves the transfer of protons (H⁺) between species, which can affect hydroxide and hydrogen ion concentrations in a solution. When discussing strong bases, these reactions are relevant in neutralization reactions, where the base reacts with an acid to form water and a salt.
For instance, if NaOH reacts with a hydrochloric acid (HCl), the reaction can be represented as:

• \(\text{NaOH} + \text{HCl} \rightarrow \text{NaCl} + \text{H}_2\text{O}\)

Even though the original exercise focused on base solutions only, understanding such interactions helps in predicting outcomes when bases are present with acids. These reactions return the \(\text{pH}\) towards neutral (pH 7) by reducing the concentration of hydroxide ions through formation of water.