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

To make a \(250.0 \mathrm{~mL}\) solution with a \(\mathrm{pH}\) of \(10.00\), \(0.0159 \mathrm{~g}\) of \(\mathrm{HONH_{2}}\) is required to dissolve in enough water.

Step-by-step solution

Calculate pOH

To find the \(\mathrm{pOH}\), use the relationship between \(\mathrm{pH}\) and \(\mathrm{pOH}\): \(\mathrm{pH} + \mathrm{pOH} = 14\) Given that the \(\mathrm{pH}\) of the solution is \(10.00\), the \(\mathrm{pOH}\) can be calculated as: \(\mathrm{pOH} = 14 - \mathrm{pH} = 14 - 10.00 = 4.00\)

Determine the concentration of OH⁻ ions

Using the \(\mathrm{pOH}\) value, we can now determine the concentration of \(\mathrm{OH}^{-}\) ions in the solution. The relationship between \(\mathrm{pOH}\) and the concentration of \(\mathrm{OH}^{-}\) ions is given by: \(\mathrm{pOH} = -\log [\mathrm{OH}^{-}]\) Solving for the concentration of \(\mathrm{OH}^{-}\) ions: \([\mathrm{OH}^{-}] = 10^{-\mathrm{pOH}} = 10^{-4.00} = 1.0 \times 10^{-4} \mathrm{M}\)

Calculate moles of HONH₂ using Kb

The base dissociation constant (\(K_{\mathrm{b}}\)) for \(\mathrm{HONH_{2}}\) is given as \(1.1 \times 10^{-8}\). The dissociation equilibrium can be written as: \(\mathrm{HONH}_{2} + \mathrm{H}_{2}\mathrm{O} \rightleftarrows \mathrm{HONH}^{+} + \mathrm{OH}^{-}\) Since we already know the concentration of \(\mathrm{OH}^{-}\) ions, and the initial concentration of \(\mathrm{HONH}^{+}\) ions can be assumed to be \(0\), we can use an ICE table to set up the equations for the equilibrium concentrations: \(I\)c\(E\)\\ \([\mathrm{HONH}_2]\) \(x-1.0 \times 10^{-4}\)\\ \([\mathrm{HONH}^{+}]\) \(0\) \(1.0 \times 10^{-4}\)\\ \([\mathrm{OH}^{-}]\) \(0\) \(1.0 \times 10^{-4}\) Substituting these concentrations into the equation for \(K_{\mathrm{b}}\): \(K_{\mathrm{b}} = \frac{[\mathrm{HONH}^{+}][\mathrm{OH}^{-}]}{[\mathrm{HONH}_2]}\) \(1.1 \times 10^{-8} = \frac{(1.0 \times 10^{-4})(1.0 \times 10^{-4})}{x-1.0 \times 10^{-4}}\) Solving for \(x\) will give us the moles of \(\mathrm{HONH_{2}}\) required: \(x = 1.818 \times 10^{-3} \mathrm{M}\)

Convert moles of HONH₂ to mass

Now that we have obtained the concentration of \(\mathrm{HONH}_2\) in the solution, we can easily determine the mass. First, we need to calculate the moles of \(\mathrm{HONH}_2\) required for \(250.0 \mathrm{~mL}\) of the solution: Moles of \(\mathrm{HONH}_{2} = (1.818 \times 10^{-3} \mathrm{M}) \times (0.250 \mathrm{~L}) = 4.545 \times 10^{-4} \mathrm{~mol}\) Next, we will find the molar mass of \(\mathrm{HONH}_2\): \(\mathrm{HONH}_2\): \(1(\mathrm{H}) + 1(\mathrm{O}) + 1(\mathrm{N}) + 2(\mathrm{H}) = 1(1.00784) + 1(15.999) + 1(14.00674) + 2(1.00784) = 35.02826 \mathrm{~g/mol}\) Finally, we will convert moles to mass: Mass of \(\mathrm{HONH}_2 = (4.545 \times 10^{-4} \mathrm{~mol}) \times (35.02826 \mathrm{~g/mol}) = 0.0159 \mathrm{~g}\) Thus, \(0.0159 \mathrm{~g}\) of \(\mathrm{HONH_{2}}\) is required to dissolve in enough water to make \(250.0 \mathrm{~mL}\) of a solution with a pH of \(10.00\).

Key concepts

pH calculation

When dealing with solutions, the pH is an essential measure. It indicates the acidity or alkalinity of a solution on a scale ranging from 0 to 14. A pH of 7 is considered neutral, below 7 acidic, and above 7 basic (alkaline). For this exercise, the given

• pH: 10.00

indicates a basic solution. To calculate the pOH, we make use of the precise relationship between pH and pOH:

\[\text{pH} + \text{pOH} = 14\]

With a pH of 10.00, the calculation of pOH becomes straightforward:\[\text{pOH} = 14 - \text{pH} = 4.00\]This result tells us that the hydroxide ion concentration is significant in this solution. Understanding the linkage between pH and pOH is vital for fostering a deeper knowledge of acid-base equilibria and preparing for subsequent calculations.

base dissociation constant

The base dissociation constant, denoted as \( K_b \), is a measure of a base's strength. Specifically, it describes the base's ability to dissociate into ions in an aqueous solution. A higher \( K_b \) value indicates a stronger base.

• Given \( K_b \) for HONH₂: \( 1.1 \times 10^{-8} \)

In this problem, the base dissociation reaction is:\[\text{HONH}_2 + \text{H}_2\text{O} \rightleftharpoons \text{HONH}^+ + \text{OH}^-\]Using an ICE (Initial, Change, Equilibrium) table, we establish a framework to determine equilibrium concentrations:- Initially, only \( \text{HONH}_2 \) is present.- Change occurs as \( \text{HONH}_2 \) dissociates to form \( \text{OH}^- \) and \( \text{HONH}^+ \).- At equilibrium, concentrations reflect initial amounts adjusted by dissociation changes.These values enable us to apply the \( K_b \) formula:\[K_b = \frac{[\text{HONH}^+][\text{OH}^-]}{[\text{HONH}_2]}\]By solving this equation, you find the concentration of unreacted \( \text{HONH}_2 \), linking these concepts back to equilibrium. This approach lets you skillfully solve for unknowns while developing an intuitive grasp of acid-base reactions.

molar mass calculation

To find the mass of a substance in a solution, the molar mass is crucial. It represents the mass of a given substance (such as \( \text{HONH}_2 \)) per mole. For this exercise, we calculate the molar mass of \( \text{HONH}_2 \) by summing the atomic masses of its constituent elements:

• Hydrogen (H): 1.00784 g/mol
• Oxygen (O): 15.999 g/mol
• Nitrogen (N): 14.00674 g/mol
• Hydrogen (additional, 2 more): 2 \times 1.00784 g/mol

The total molar mass of \( \text{HONH}_2 \) is calculated as:\[1(1.00784) + 1(15.999) + 1(14.00674) + 2(1.00784) = 35.02826 \text{ g/mol}\]Once the molar mass is determined, it becomes possible to convert the moles of \( \text{HONH}_2 \) into grams for any given molarity and volume, by using\[\text{Mass} = \text{Moles} \times \text{Molar Mass}\]In this case, with a solution volume of 0.250 L and a requisite mole concentration, calculating the precise mass accurately facilitates preparation and experimental success, demonstrating thorough application of foundational chemistry concepts.