Problem 11
Question
Calculate \(\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]\) and \(\mathrm{pH}\) in saturated \(\mathrm{Ba}(\mathrm{OH})_{2}(\mathrm{aq})\) which contains \(3.9 \mathrm{g} \mathrm{Ba}(\mathrm{OH})_{2} \cdot 8 \mathrm{H}_{2} \mathrm{O}\) per \(100 \mathrm{mL}\) of solution.
Step-by-Step Solution
Verified Answer
The concentration of the \(\mathrm{H}_{3} \mathrm{O}^{+}\) ions is \(4.03 \times 10^{-16} \: M\) and the pH of the solution is 15.4.
1Step 1: Calculate number of moles
Calculate the number of moles of \(\mathrm{Ba}(\mathrm{OH})_{2}(\mathrm{aq})\) in 100 mL of the solution using the molar mass (formula weight \(\mathrm{Ba}(\mathrm{OH})_{2} \cdot 8 \mathrm{H}_{2} O = 315.46 \: g/mol\)). You do this by dividing the mass of solute by the molar mass. \(n = \frac{3.9 \: g}{315.46 \: g/mol} = 0.0124 \: mol\).
2Step 2: Calculate concentration of hydroxide ions
Calculate the concentration of \(\mathrm{OH}^-\) ions in the solution. Since each formula unit of \(\mathrm{Ba}(\mathrm{OH})_{2}\) contains 2 \(\mathrm{OH}^-\) ions, the concentration of \(\mathrm{OH}^-\) ions will be twice the number of moles of \(\mathrm{Ba}(\mathrm{OH})_{2}\) since it dissociates fully in solution in 1L, but since it's in 100mL the concentration \([\mathrm{OH}^-]\) will be \(\frac{(2*0.0124mol)}{100 mL} * 1000 = 0.248 M\)
3Step 3: Calculate concentration of hydronium ions
Use the ion-product of water (\(K_w = [\mathrm{H}_{3} \mathrm{O}^{+}][\mathrm{OH}^{-}]\) = \(1.0 \times 10^{-14}\)) to find concentration of \(\mathrm{H}_{3} \mathrm{O}^{+}\) ions. Rearrange the formula to find \([\mathrm{H}_{3} \mathrm{O}^{+}]\), which equals to \(K_w / [\mathrm{OH}^{-}] = \frac{1.0 \times 10^{-14}}{0.248} = 4.03 \times 10^{-16} M\).
4Step 4: Calculate pH
Finally find the pH using the formula pH = -\(\log[\mathrm{H}_{3} \mathrm{O}^{+}]\), so the pH = \( -\log(4.03 \times 10^{-16}) = 15.4\).
Key Concepts
pH CalculationMolar ConcentrationIon-Product of Water
pH Calculation
Understanding how to calculate the pH of a solution is an essential skill in chemistry. pH is a measure of the acidity or basicity of an aqueous solution. On the pH scale, which ranges from 0 to 14, solutions with a pH less than 7 are acidic, and solutions with a pH greater than 7 are basic.
The pH is calculated using the concentration of hydronium ions \(\mathrm{H}_{3}\mathrm{O}^{+}\) in the solution. The relationship between pH and hydronium ion concentration is given by the formula:
\[\text{pH} = -\log\left([\mathrm{H}_{3}\mathrm{O}^{+}]\right)\]
This means to find the pH, you take the negative logarithm (base 10) of the molar concentration of hydronium ions. A key point to remember is that a higher concentration of hydronium ions corresponds to a lower pH (more acidic), and a lower concentration corresponds to a higher pH (more basic).
The pH is calculated using the concentration of hydronium ions \(\mathrm{H}_{3}\mathrm{O}^{+}\) in the solution. The relationship between pH and hydronium ion concentration is given by the formula:
\[\text{pH} = -\log\left([\mathrm{H}_{3}\mathrm{O}^{+}]\right)\]
This means to find the pH, you take the negative logarithm (base 10) of the molar concentration of hydronium ions. A key point to remember is that a higher concentration of hydronium ions corresponds to a lower pH (more acidic), and a lower concentration corresponds to a higher pH (more basic).
Molar Concentration
Molar concentration, often simply called concentration, refers to the amount of a substance (solute) dissolved in a given volume of solvent. It's denoted by the unit molarity (M), which is moles per liter (mol/L).
To calculate the molar concentration, you'll need to:
To calculate the molar concentration, you'll need to:
- Determine the number of moles of solute present.
- Measure the volume of the solution in liters.
- Divide the moles of solute by the volume of the solution.
Ion-Product of Water
The ion-product of water \(K_w\) is a constant value that represents the product of the concentrations of hydronium \(\mathrm{H}_{3}\mathrm{O}^{+}\) and hydroxide \(\mathrm{OH}^{-}\) ions in water at a given temperature. At room temperature, \(K_w\) is \(1.0 \times 10^{-14}\).\[K_w = [\mathrm{H}_{3} \mathrm{O}^{+}][\mathrm{OH}^{-}] = 1.0 \times 10^{-14}\]
This constant is an important concept when solving for unknown concentrations in acid-base chemistry. If you know the concentration of one of the ions, you can easily solve for the other using this constant. This relationship is what allows us to find the \(\mathrm{H}_{3}\mathrm{O}^{+}\) concentration if the \(\mathrm{OH}^{-}\) concentration is known, providing a link between pH and the basicity of a solution.
This constant is an important concept when solving for unknown concentrations in acid-base chemistry. If you know the concentration of one of the ions, you can easily solve for the other using this constant. This relationship is what allows us to find the \(\mathrm{H}_{3}\mathrm{O}^{+}\) concentration if the \(\mathrm{OH}^{-}\) concentration is known, providing a link between pH and the basicity of a solution.
Other exercises in this chapter
Problem 9
Calculate \(\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]\) and \(\left[\mathrm{OH}^{-}\right]\) for each solution:(a) \(0.00165 \mathrm{M} \mathrm{HNO}_{3} ;\) (b
View solution Problem 10
What is the pH of each of the following solutions? (a) \(0.0045 \mathrm{M} \mathrm{HCl} ;\) (b) \(6.14 \times 10^{-4} \mathrm{M} \mathrm{HNO}_{3} ;\) (c) 0.0068
View solution Problem 12
A saturated aqueous solution of \(\mathrm{Ca}(\mathrm{OH})_{2}\) has a pH of 12.35. What is the solubility of \(\mathrm{Ca}(\mathrm{OH})_{2}\), expressed in mil
View solution Problem 14
What is the \(\mathrm{pH}\) of the solution obtained when \(125 \mathrm{mL}\) of \(0.606 \mathrm{M} \mathrm{NaOH}\) is diluted to \(15.0 \mathrm{L}\) with water
View solution