Question

Calculate the mass of \(\mathrm{HONH}_{2}\) required to dissolve in enough water to make \(250.0 \mathrm{~mL}\) of solution having a \(\mathrm{pH}\) of \(10.00\left(K_{\mathrm{b}}=\right.\) \(\left.1.1 \times 10^{-8}\right)\)

Short answer

Approximately 0.0075 g of HONH2 is required to dissolve in enough water to make 250.0 mL of solution having a pH of 10.00.

Step-by-step solution

Convert pH to pOH

Since we know \(pH + pOH = 14\), we can find the pOH by subtracting the given pH from 14: \(pOH = 14 - 10 = 4\)

Find the concentration of OH- ions

To find the concentration of OH- ions, we use the relationship between pOH and [OH-]: \(pOH = -\log[OH^{-}]\) Now we can solve for [OH-]: \([OH^{-}] = 10^{-pOH} = 10^{-4}\)

Use the concentration of OH- ions and Kb to find the concentration of HONH2

We use the expression for Kb: \(K_b = \frac{[OH^{-}][HONH^{+}]}{[HONH_{2}]}\) Since HONH2 is a weak base, we assume that initially, none of it has dissociated into ions. Then, as a small amount x dissociates, we can write: \[K_b = \frac{x * x}{(M_{initial} - x)}\] where - x is the concentration of dissociated HONH2 - M_initial is the initial concentration of HONH2 We know the \(K_b = 1.1 \times 10^{-8}\) and \([OH^{-}] = x = 10^{-4}\). Substituting these values in the equation: \(1.1 \times 10^{-8} = \frac{(10^{-4})^{2}}{(M_{initial} - 10^{-4})}\)

Solve for the initial concentration of HONH2

Rearrange the equation and solve for M_initial: \(M_{initial} = \frac{(10^{-4})^{2}}{1.1 \times 10^{-8}} + 10^{-4} \approx 9.09 \times 10^{-4}\mathrm{~M}\)

Calculate moles and mass of HONH2 needed

We know the volume of the solution required is 250.0 mL (or 0.250 L). We can now find the moles of HONH2 needed: Moles = M_initial * Volume = 9.09 × 10^(-4) M * 0.250 L = 2.27 × 10^(-4) moles The molar mass of HONH2 is approximately 33 g/mol. Thus, we can find the mass required: Mass = Moles * Molar mass = 2.27 × 10^(-4) moles * 33 g/mol ≈ 0.0075 g So, approximately 0.0075 g of HONH2 is required to dissolve in enough water to make 250.0 mL of solution having a pH of 10.00.

Key concepts

pOH and pH Relationship

Understanding the relationship between pOH and pH is crucial when dealing with solutions in chemistry. The pH scale measures how acidic or basic a solution is. It ranges from 0 to 14, with 0 being very acidic, 7 neutral, and 14 very basic. The pOH scale, on the other hand, directly relates to the concentration of hydroxide ions (OH-). The pOH value is found by taking the negative logarithm of the hydroxide ion concentration.

The pH and pOH scales are inversely related — when one increases, the other decreases. This is expressed by the equation:
\[ pH + pOH = 14 \] This formula is a crucial aspect of pH calculation chemistry and allows us to convert between pH and pOH, giving us a complete picture of the solution's acidity or basicity. For instance, if a solution has a pH of 10, like in our exercise, the pOH would be 4. With this information, we can calculate the concentration of hydroxide ions in the solution.

Hydroxide Ion Concentration

The hydroxide ion concentration in a solution is a direct measure of its basicity. More hydroxide ions mean a more basic solution. To determine the concentration of these ions, we use the pOH value obtained from the pH, as explained previously.

The relationship between pOH and hydroxide ion concentration is given by the equation:
\[ pOH = -\log[OH^{-}] \] Solving for the hydroxide ion concentration, we get:
\[ [OH^{-}] = 10^{-pOH} \] In our exercise, with a pOH of 4, the hydroxide ion concentration is \(10^{-4}\text{M}\). This value is essential for calculating the amount of weak base that can dissociate in a solution.

Weak Base Dissociation

Weak bases do not completely dissociate in water, unlike strong bases. During dissociation, a weak base like HONH2 partially separates into its constituent ions. To describe this process quantitatively, we use the base dissociation constant, Kb, which provides a measure of how readily a weak base dissociates.

The formula for Kb is:
\[ K_b = \frac{[OH^{-}][HONH^{+}]}{[HONH_{2}]} \] For a weak base undergoing dissociation, we generally assume that the concentration of dissociated ions is equal to 'x', allowing us to rewrite the formula as:
\[ K_b = \frac{x \cdot x}{(M_{initial} - x)} \] Here, 'x' represents the concentration of hydroxide ions produced, and 'M_initial' stands for the initial concentration of the weak base. In our exercise, we were given Kb and the concentration of OH-, which allowed us to calculate the initial concentration of the weak base.

Molarity and Moles Calculation

Molarity is a standard measure of concentration in chemistry, defined as moles of solute per liter of solution (\text{M}\ = \frac{\text{moles}}{\text{liter}}\text{)}). To find the mass of a compound required to achieve a specific molarity, we first calculate the number of moles needed. Then, using the compound's molar mass, we calculate the mass.

For the exercise, with the initial molarity (M_initial) and the volume of the solution, we calculated the moles of HONH2 as:
\[ \text{Moles} = M_{initial} \times \text{Volume} \] Finally, the mass of HONH2 required was computed by:
\[ \text{Mass} = \text{Moles} \times \text{Molar mass} \] By using these calculations, we determined the exact amount of HONH2 necessary to create a solution with the desired pH.