Question

A solution is made by mixing \(15.0 \mathrm{~g}\) of \(\mathrm{Sr}(\mathrm{OH})_{2}\) and \(55.0 \mathrm{~mL}\) of \(0.200 \mathrm{MHNO}_{3}\). (a) Write a balanced equation for the reaction that occurs between the solutes. (b) Calculate the concentration of each ion remaining in solution. (c) Is the resultant solution acidic or basic?

Short answer

The balanced chemical equation for the reaction between strontium hydroxide \(\mathrm{Sr(OH)_2}\) and nitric acid \(\mathrm{HNO_3}\) is: \[\mathrm{Sr(OH)_2 + 2HNO_3 \rightarrow Sr(NO_3)_2 + 2H_2O}\] The concentration of each ion remaining in the solution is: \[\mathrm{Sr^{2+}: 2.24 \,M}\] \[\mathrm{NO_3^-: 0.400\,M}\] \[\mathrm{OH^-: 0\,M}\] The resultant solution is acidic since there are no \(\mathrm{OH^-}\) ions left in the solution, and unreacted \(\mathrm{HNO_3}\) is present.

Step-by-step solution

(a) Balanced Chemical Equation

First, let's write the chemical equation for the reaction between strontium hydroxide \(\mathrm{Sr}(OH)_2\) and nitric acid \(\mathrm{HNO}_3\). \(\mathrm{Sr(OH)_2 + 2HNO_3 \rightarrow Sr(NO_3)_2 + 2H_2O}\) This is the balanced chemical equation for this reaction.

(b) Concentration of Each Ion Remaining in Solution

Now we need to calculate the moles of the reactants present in the solution, compare them and determine the limiting reactant and the remaining ions. 1. Calculate moles of \(\mathrm{Sr}(OH)_2\): Given mass of \(\mathrm{Sr}(OH)_2 = 15.0\,\text{g}\) The molar mass of \(\mathrm{Sr}(OH)_2 = 121.6\, \mathrm{g/mol}\) Moles of \(\mathrm{Sr}(OH)_2 = \frac{\text{mass of Sr(OH)_2}}{\text{molar mass of Sr(OH)_2}}\) Moles of \(\mathrm{Sr}(OH)_2 = \frac{15.0\,\text{g}}{121.6 \,\mathrm{g/mol}} = 0.123\, \mathrm{mol}\) 2. Calculate moles of \(\mathrm{HNO}_3\): Given volume of \(\mathrm{HNO}_3 = 55.0 \,\mathrm{mL}\) and its concentration \(= 0.200\,\mathrm{M}\) Moles of \(\mathrm{HNO}_3 = \text{concentration} \times \text{volume in L}\) Moles of \(\mathrm{HNO}_3 = 0.200 \,\text{M} \times 0.055\,\text{L} = 0.0110\,\text{mol}\) 3. Determine the limiting reactant: The balanced equation shows that we need two moles of \(\mathrm{HNO}_3\) for every one mole of \(\mathrm{Sr}(OH)_2\). So we need to compare the moles available for each reactant: \(\mathrm{\frac{0.0110\,mol\, of \ HNO_3} {0.123\,mol\, of\, Sr(OH)_2}} = 0.0894\, \mathrm{mol}\) Since \(0.0894 \mathrm{mol} < 1 \mathrm{mol}\), \(\mathrm{HNO}_3\) is the limiting reactant. Thus, additional \(0.200\,\mathrm{M}\,\mathrm{HNO}_3\) is needed to react completely with \(\mathrm{Sr}(OH)_2\). 4. Calculate the concentration of \(\mathrm{Sr^{2+}}\) and \(\mathrm{NO_3^-}\): For \(\mathrm{Sr^{2+}}\), we can directly calculate: Concentration of \(\mathrm{Sr^{2+}} = \frac{\text{moles of Sr(OH)_2}}{\text{total volume in L}}\) Concentration of \(\mathrm{Sr^{2+}} = \frac{0.123\,\text{mol}}{0.055\,\text{L}} = 2.24\,\mathrm{M}\) For \(\mathrm{NO_3^-}\), we need to add the moles that came from remaining \(\mathrm{HNO}_3\): Moles of $\mathrm{NO_3^- from HNO_3} = 2 \times \text{moles of }\mathrm{HNO_3} = 2 \times 0.0110 \,\text{mol} = 0.0220\,\text{mol}$ Concentration of \(\mathrm{NO_3^-} = \frac{\text{moles of NO_3^-}}{\text{total volume in L}}\) Concentration of \(\mathrm{NO_3^-} = \frac{0.0220\,\text{mol}}{0.055\,\text{L}} = 0.400\,\mathrm{M}\) 5. Calculate the concentration of remaining \(\mathrm{OH^-}\): Since both moles of \(\mathrm{HNO_3}\) reacted with \(\mathrm{Sr}(OH)_2\), no \(\mathrm{OH^-}\) ions are left in the solution. So, concentration of each ion in the solution is: \(\mathrm{Sr^{2+}}: 2.24 \,\mathrm{M}\) \(\mathrm{NO_3^-}: 0.400\,\mathrm{M}\) \(\mathrm{OH^-}: 0\,\mathrm{M}\)

(c) Determine if the Resultant Solution is Acidic or Basic

Since there are no \(\mathrm{OH^-}\) ions left in the solution, the final solution is acidic due to the presence of the unreacted \(\mathrm{HNO}_3\).

Key concepts

Balancing Chemical Equations

Understanding how to balance chemical equations is pivotal when studying chemistry. A chemical equation represents the reactants and products in a chemical reaction through symbols and formulas. For an equation to be balanced, the number of atoms for each element must be equal on both sides of the reaction.

For example, in the reaction between strontium hydroxide \(\mathrm{Sr}(OH)_2\) and nitric acid \(\mathrm{HNO}_3\), we must balance the equation to comply with the law of conservation of mass. The balanced equation is \(\mathrm{Sr(OH)_2 + 2HNO_3 \rightarrow Sr(NO_3)_2 + 2H_2O}\). Here, we can see that there are equal numbers of each type of atom on both sides of the equation: one strontium (Sr), two nitrogen (N), six oxygen (O), and two hydrogen (H) atoms.

When balancing equations, it's helpful to start by balancing the elements that appear in the smallest amount and are not diatomic molecules (such as O\(_2\) and H\(_2\)), and then proceed to balance the more complex ones. If an element appears in more than one compound on a side of the equation, balance it last.

Calculating Concentration of Ions

Calculating the concentration of ions in a solution is an essential skill in chemistry, especially in fields like analytical chemistry. Concentration can be expressed in terms of molarity (M), which is moles of solute per liter of solution.

To calculate the concentration, you first need the number of moles of the ion of interest. For instance, in the example given, we start with 15.0 grams of strontium hydroxide \(\mathrm{Sr}(OH)_2\) and determine the moles of \(\mathrm{Sr}^{2+}\) ions by dividing by the compound's molar mass. Once the moles of \(\mathrm{Sr}^{2+}\) are known, you can find the concentration by dividing the moles by the total volume of the solution in liters. The process is similar for \(\mathrm{NO_3^-}\) ions, with an added step of considering reactions that may affect the ion's availability, such as the remaining \(\mathrm{HNO}_3\).

For complex solutions, where reactions are taking place, it's crucial to consider the stoichiometry of the reaction to know how reactants and products relate, thus altering concentrations. Always remember that the total volume of the solution might change after a reaction, especially if additional substances are added or if there is significant volume contraction.

Determining Solution Acidity or Basicity

The acidity or basicity of a solution is an indicator of its pH level, which is a measure of the hydrogen ion concentration \(\mathrm{[H^+]}\). Acidity and basicity are commonly determined based on the presence of excess hydronium \(\mathrm{H_3O^+}\) or hydroxide \(\mathrm{OH^-}\) ions. An acidic solution has more hydronium ions than hydroxide ions, while a basic solution has more hydroxide ions.

In the example provided, after the reaction between strontium hydroxide and nitric acid, no hydroxide ions remain in the solution, indicating that all of them have been neutralized. The presence of unreacted nitric acid suggests that the solution is acidic. To determine the exact pH, one might need to use a pH meter or an acid-base indicator.

The pH scale ranges from 0 (most acidic) to 14 (most basic), with 7 being neutral. It is logarithmic, meaning each whole pH value below 7 is ten times more acidic than the next higher value, and likewise, each whole pH value above 7 is ten times less acidic (or more basic) than the next lower value. Understanding how to calculate the effects of different ions on the solution's pH is critical for predicting the behavior of the solution in different contexts, like chemical reactions or biological systems.