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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

Expert verified
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

01

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: \(\left[\mathrm{OH}^{-}\right]\) = \(2(1.5 \times 10^{-3}) = 3.0 \times 10^{-3} \text{M}\)
02

Calculate the pOH

Use the equation \(\mathrm{pOH} = - \log_{10}[\mathrm{OH}^{-}]\) to calculate the pOH: \(\mathrm{pOH} = - \log_{10}(3.0 \times 10^{-3}) = 2.52\)
03

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
04

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 = \(\frac{2.250\,\text{g}}{6.94\,\text{g/mol} + 16.00\,\text{g/mol} + 1.01\,\text{g/mol}}=\frac{2.250\,\text{g}}{23.95\,\text{g/mol}}\approx 0.0940\,\text{moles}\)
05

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.
06

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: \(\left[\mathrm{OH}^{-}\right] = \frac{0.0940\,\text{moles}}{0.250\,\text{L}} = 0.376\,\text{M}\)
07

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
08

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
09

Calculate the concentration of OH- ions

Divide the moles of OH- ions by the total volume (in Liters) to find the hydroxide ion concentration: \(\left[\mathrm{OH}^{-}\right] = \frac{1.75\times10^{-4}\,\text{moles}}{2.00\,\text{L}} = 8.75\times10^{-5}\,\text{M}\)
10

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 \(9.5 \times 10^{-2} \mathrm{M Ca(OH)_{2}}\)
11

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
12

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
13

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: \(\left[\mathrm{OH}^{-}\right] = \frac{(5.25\times10^{-4}\,\text{moles} + 2.85\times10^{-3}\,\text{moles})}{0.020\,\text{L}} \approx 0.158\,\mathrm{M}\)
14

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 \(\mathrm{pH} = 13.199\)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.

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Most popular questions from this chapter

Calculate the \(\mathrm{pH}\) of each of the following strong acid solutions: \((\mathbf{a}) 8.5 \times 10^{-3} \mathrm{M} \mathrm{HBr},(\mathbf{b}) 1.52 \mathrm{g}\) of \(\mathrm{HNO}_{3}\) in 575 \(\mathrm{mL}\) of solution, \((\mathbf{c}) 5.00 \mathrm{mL}\) of 0.250 \(\mathrm{M} \mathrm{ClO}_{4}\) diluted to 50.0 \(\mathrm{mL}\) (d) a solution formed by mixing 10.0 \(\mathrm{mL}\) of 0.100 \(\mathrm{M} \mathrm{HBr}\) with 20.0 \(\mathrm{mL}\) of 0.200 \(\mathrm{M} \mathrm{HCl} .\)

Indicate whether each of the following statements is correct or incorrect. (a) Every Bronsted-Lowry acid is also a Lewis acid. (b) Every Lewis acid is also a Bronsted-Lowry acid. (c) Conjugate acids of weak bases produce more acidic solutions than conjugate acids of strong bases. (d) \(\mathrm{K}^{+}\) ion is acidic in water because it causes hydrating water molecules to become more acidic. (e) The percent ionization of a weak acid in water increases as the concentration of acid decreases.

By what factor does \(\left[\mathrm{H}^{+}\right]\) change for a pH change of \((\mathbf{a}) 2.00\) units, \((\mathbf{b}) 0.50\) units?

What is the boiling point of a 0.10\(M\) solution of \(\mathrm{NaHSO}_{4}\) if the solution has a density of 1.002 \(\mathrm{g} / \mathrm{mL} ?\)

Indicate whether each of the following statements is true or false. For each statement that is false, correct the statement to make it true. (a) In general, the acidity of binary acids increases from left to right in a given row of the periodic table. (b) In a series of acids that have the same central atom, acid strength increases with the number of hydrogen atoms bonded to the central atom. (c) Hydrotelluric acid \(\left(\mathrm{H}_{2} \mathrm{Te}\right)\) is a stronger acid than \(\mathrm{H}_{2} \mathrm{S}\) because Te is more electronegative than \(\mathrm{S} .\)

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