Question

Calculate \(\left[\mathrm{OH}^{-}\right]\) and \(\mathrm{pH}\) for (a) \(1.5 \times 10^{-3} \mathrm{MSr}(\mathrm{OH})_{2}\) (b) 2.250 \(\mathrm{g}\) of LiOH in 250.0 \(\mathrm{mL}\) of solution, \((\mathbf{c}) 1.00\) mL of 0.175 M NaOH 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 \(9.5 \times 10^{-2} \mathrm{MCa}(\mathrm{OH})_{2}.\)

Short answer

In summary, the solutions for the given exercise are: (a) \(\left[\mathrm{OH}^{-}\right] = 3.0 \times 10^{-3} \mathrm{M}\) and \(\mathrm{pH} = 11.48\) (b) \(\left[\mathrm{OH}^{-}\right] = 0.376 \mathrm{M}\) and \(\mathrm{pH} = 13.576\) (c) \(\left[\mathrm{OH}^{-}\right] = 8.75\times10^{-5} \mathrm{M}\) and \(\mathrm{pH} = 9.942\) (d) \(\left[\mathrm{OH}^{-}\right] = 0.158\mathrm{M}\) and \(\mathrm{pH} = 13.199\)

Step-by-step solution

Determine the number of moles of OH- ions

Each mole of Sr(OH)\(_{2}\) will dissociate into 2 moles of OH- ions. Therefore, multiply the concentration by two to find the concentration of hydroxide ions: \(\left[\mathrm{OH}^{-}\right]\) = \(2(1.5 \times 10^{-3}) = 3.0 \times 10^{-3} \text{M}\)

Calculate the pOH

Use the equation \(\mathrm{pOH} = - \log_{10}[\mathrm{OH}^{-}]\) to calculate the pOH: \(\mathrm{pOH} = - \log_{10}(3.0 \times 10^{-3}) = 2.52\)

Calculate the pH

Lastly, use the equation \(\mathrm{pH} = 14 - \mathrm{pOH}\) to find the pH: \(\mathrm{pH} = 14 - 2.52 = 11.48\) Part (a) Results: \(\left[\mathrm{OH}^{-}\right] = 3.0 \times 10^{-3} \mathrm{M}\) and \(\mathrm{pH} = 11.48\) Part (b): 2.250 g of LiOH in 250.0 mL of solution

Calculate mole of LiOH

We first need to convert the mass of LiOH to moles of LiOH using its molar mass (Li: 6.94 g/mol, O: 16.00 g/mol, H: 1.01 g/mol): Moles of LiOH = \(\frac{2.250\,\text{g}}{6.94\,\text{g/mol} + 16.00\,\text{g/mol} + 1.01\,\text{g/mol}}=\frac{2.250\,\text{g}}{23.95\,\text{g/mol}}\approx 0.0940\,\text{moles}\)

Determine the moles of OH- ions

LiOH dissociates into Li+ and OH- ions in a 1:1 ratio. Therefore, the moles of OH- ions will be equal to the moles of LiOH.

Calculate the concentration of OH- ions

Divide the moles of OH- ions by the volume of the solution (in Liters) to find the concentration of hydroxide ions: \(\left[\mathrm{OH}^{-}\right] = \frac{0.0940\,\text{moles}}{0.250\,\text{L}} = 0.376\,\text{M}\)

Calculate the pOH and pH

Using the same equations as before, we find the pOH: \(\mathrm{pOH} = - \log_{10}(0.376) \approx 0.424\) And then find the pH: \(\mathrm{pH} = 14 - 0.424 \approx 13.576\) Part (b) Results: \(\left[\mathrm{OH}^{-}\right] = 0.376 \mathrm{M}\) and \(\mathrm{pH} = 13.576\) Part (c): 1.00 mL of 0.175 M NaOH diluted to 2.00 L

Calculate moles of OH- ions

Calculate the moles of OH- ions from the initial concentration: Moles of OH- ions = (0.175 mol/L)(0.001 L) = 1.75 x 10^(-4) mol

Calculate the concentration of OH- ions

Divide the moles of OH- ions by the total volume (in Liters) to find the hydroxide ion concentration: \(\left[\mathrm{OH}^{-}\right] = \frac{1.75\times10^{-4}\,\text{moles}}{2.00\,\text{L}} = 8.75\times10^{-5}\,\text{M}\)

Calculate the pOH and pH

Using the equations as before, find the pOH: \(\mathrm{pOH} = - \log_{10}(8.75\times10^{-5}) \approx 4.058\) And then find the pH: \(\mathrm{pH} = 14 - 4.058 = 9.942\) Part (c) Results: \(\left[\mathrm{OH}^{-}\right] = 8.75\times10^{-5} \mathrm{M}\) and \(\mathrm{pH} = 9.942\) Part (d): a solution formed by adding 5.00 ml of 0.105 M KOH to 15.0 ml of \(9.5 \times 10^{-2} \mathrm{M Ca(OH)_{2}}\)

Calculate moles of OH- ions in each solution

Calculate the total moles of OH- ions from both the KOH and Ca(OH)\(_{2}\) solutions: Moles of OH- ions from KOH = (0.105 mol/L)(0.005 L) = 5.25 x 10^(-4) moles Moles of OH- ions from Ca(OH)\(_{2}\) = 2(9.5 x 10^(-2) mol/L)(0.015 L) = 2.85 x 10^(-3) moles

Calculate the total volume

Add together the volumes of both solutions: Total Volume = 5.00 mL + 15.0 mL = 20.0 mL = 0.020 L

Calculate the concentration of OH- ions

Add the moles of OH- ions from both solutions, and then divide by the total volume to find the hydroxide ion concentration: \(\left[\mathrm{OH}^{-}\right] = \frac{(5.25\times10^{-4}\,\text{moles} + 2.85\times10^{-3}\,\text{moles})}{0.020\,\text{L}} \approx 0.158\,\mathrm{M}\)

Calculate the pOH and pH

Using the equations as before, find the pOH: \(\mathrm{pOH} = - \log_{10}(0.158) \approx 0.801\) And then find the pH: \(\mathrm{pH} = 14 - 0.801 = 13.199\) Part (d) Results: \(\left[\mathrm{OH}^{-}\right] = 0.158\mathrm{M}\) and \(\mathrm{pH} = 13.199\) In conclusion, the solutions for the given exercise are as follows: (a) \(\left[\mathrm{OH}^{-}\right] = 3.0 \times 10^{-3} \mathrm{M}\) and \(\mathrm{pH} = 11.48\) (b) \(\left[\mathrm{OH}^{-}\right] = 0.376 \mathrm{M}\) and \(\mathrm{pH} = 13.576\) (c) \(\left[\mathrm{OH}^{-}\right] = 8.75\times10^{-5} \mathrm{M}\) and \(\mathrm{pH} = 9.942\) (d) \(\left[\mathrm{OH}^{-}\right] = 0.158\mathrm{M}\) and \(\mathrm{pH} = 13.199\)

Key concepts

pH calculation

The pH scale is a measure used to determine the acidity or basicity of a solution. Calculating the pH involves understanding its relationship with pOH, the power of hydroxide ions concentration. The core formula linking them is: \[ \text{pH} + \text{pOH} = 14 \] Consequently, once you know the pOH, you can easily find the pH by subtracting the pOH from 14. For instance, part (a) of the problem demonstrated this: after computing a pOH of 2.52, the pH is simply \( 14 - 2.52 = 11.48 \). The pH value helps classify solutions as acidic (pH < 7), neutral (pH = 7), or basic (pH > 7). With substances like Sr(OH)\(_2\), LiOH, and NaOH, you would expect basic pH values as they are known bases.

hydroxide ion concentration

Hydroxide ion concentration, denoted by \( [\text{OH}^{-}] \), is a critical factor in determining the basicity of a solution. To find it, first consider the dissociation of the base in water. For example, in Sr(OH)\(_2\), each mole provides two moles of \( \text{OH}^{-} \) ions. Therefore, you multiply the molarity of Sr(OH)\(_2\) by 2 for the \( [\text{OH}^{-}] \).In the exercise:- Begin with a base's molarity or moles (calculated from mass).- Remember dissociation ratios: Sr(OH)\(_2\) provides two \( \text{OH}^{-} \) ions per unit, while LiOH offers one.Calculated \( [\text{OH}^{-}] \) directly links to pOH, using the formula:\[ \text{pOH} = -\log_{10}([\text{OH}^{-}]) \] This highlights how greater \( [\text{OH}^{-}] \) signifies a more substantial base tendency, visibly lowering the pOH and, in turn, raising the pH.

dilution calculations

Dilution refers to reducing the concentration of a solution by increasing its volume, often by adding more solvent. To find the new concentration after dilution, apply:\[ C_1V_1 = C_2V_2 \] where \( C_1 \) and \( C_2 \) are the initial and final concentrations, and \( V_1 \) and \( V_2 \) are the initial and final volumes.For example, in part (c), when diluting 1.00 mL of 0.175 M NaOH to 2.00 L, solve:\[ (0.175)(0.001) = (C_2)(2.00) \] Resulting in:\[ C_2 = \frac{0.175 \times 0.001}{2.00} = 8.75 \times 10^{-5} \text{ M} \]This lowering of concentration means fewer \( \text{OH}^{-} \) ions per unit volume, reflecting on the pH and solution properties. Understanding dilution is crucial for accurately depicting a solution's strength post-adjustment.