Question

Write a balanced net ionic equation for the reaction of each of the following aqueous solutions with \(\mathrm{H}^{+}\) ions. (a) sodium fluoride (b) barium hydroxide (c) potassium dihydrogen phosphate \(\left(\mathrm{KH}_{2} \mathrm{PO}_{4}\right)\)

Short answer

Question: Write the balanced net ionic equations for the reaction of each of the following aqueous solutions with H+ ions: (a) sodium fluoride, (b) barium hydroxide, and (c) potassium dihydrogen phosphate. Answer: (a) \(F^-(aq) + H^+(aq) \rightarrow HF(aq)\) (b) \(Ba^{2+}(aq) + 2OH^-(aq) + 4H^+(aq) \rightarrow Ba^{2+}(aq) + 4H_2O(l)\) (c) \(H_2PO_4^-(aq) + H^+(aq) \rightarrow H_3PO_4(aq)\)

Step-by-step solution

Write the balanced molecular equations

For each aqueous solution, write a balanced molecular equation for its reaction with H+ ions: (a) Sodium Fluoride: \(\mathrm{NaF(aq) + H_2SO_4(aq) \rightarrow NaHSO_4(aq) + HF(aq)}\) (b) Barium Hydroxide: \(\mathrm{Ba(OH)_2(aq) + 2H_2SO_4(aq) \rightarrow BaSO_4(s) + 4H_2O(l)}\) (c) Potassium Dihydrogen Phosphate: \(\mathrm{KH_2PO_4(aq) + H_2SO_4(aq) \rightarrow KHSO_4(aq) + H_3PO_4(aq)}\)

Write the complete ionic equations

Rewrite each balanced molecular equation as a complete ionic equation: (a) Sodium Fluoride: \(\mathrm{Na^+(aq) + F^-(aq) + 2H^+(aq) + SO_4^{2-}(aq) \rightarrow Na^+(aq) + HSO_4^-(aq) + HF(aq)}\) (b) Barium Hydroxide: \(\mathrm{Ba^{2+}(aq) + 2OH^-(aq) + 4H^+(aq) + 2SO_4^{2-}(aq) \rightarrow BaSO_4(s) + 4H_2O(l)}\) (c) Potassium Dihydrogen Phosphate: \(\mathrm{K^+(aq) + H_2PO_4^-(aq) + 2H^+(aq) + SO_4^{2-}(aq) \rightarrow K^+(aq) + HSO_4^-(aq) + H_3PO_4(aq)}\)

Write the balanced net ionic equations

Eliminate the spectator ions and write the balanced net ionic equations for each reaction: (a) Sodium Fluoride: \(\mathrm{F^-(aq) + H^+(aq) \rightarrow HF(aq)}\) (b) Barium Hydroxide: \(\mathrm{Ba^{2+}(aq) + 2OH^-(aq) + 4H^+(aq) \rightarrow Ba^{2+}(aq) + 4H_2O(l)}\) (c) Potassium Dihydrogen Phosphate: \(\mathrm{H_2PO_4^-(aq) + H^+(aq) \rightarrow H_3PO_4(aq)}\)

Key concepts

acid-base reactions

Acid-base reactions are fundamental interactions where an acid donates a proton (\(\mathrm{H^+}\)) to a base. This leads to the formation of water and a salt. For example, in the reaction with barium hydroxide and sulfuric acid, \(\mathrm{Ba(OH)_2}\) and \(\mathrm{H_2SO_4}\), neutralization occurs. The acid (\(\mathrm{H_2SO_4}\)) provides protons, which are accepted by the base (\(\mathrm{Ba(OH)_2}\)), resulting in water and insoluble barium sulfate.

These reactions are important because they allow us to understand how substances interact at a molecular level. Commonly, they are used in laboratory settings to determine concentrations and check the strength of acids or bases. To simplify, always remember:

• Acid donates \(\mathrm{H^+}\) ions.
• Base receives \(\mathrm{OH^-}\) ions.
• Resulting product often includes water and a salt.

spectator ions

Spectator ions are ions that exist in the same form on both the reactant and product sides of a chemical reaction equation. They do not participate in the reaction and can be removed when writing the net ionic equation. For example, in the reaction between sodium fluoride and sulfuric acid, sodium (\(\mathrm{Na^+}\)) and hydrogen sulfate (\(\mathrm{HSO_4^-}\)) ions appear unchanged on both sides.

They "watch" the reaction without participating. Identifying these ions helps simplify chemical equations by focusing only on the particles that undergo change. In net ionic equations, we eliminate these spectator ions, highlighting the true chemical change happening.

balanced chemical equations

Balanced chemical equations ensure that the same number of each type of atom is present on both sides of the equation. This is essential because the law of conservation of mass tells us that matter cannot be created or destroyed in chemical reactions. Balancing ensures stoichiometric equivalence between reactants and products, reflecting real-world reactions.

For example, consider the reaction between potassium dihydrogen phosphate \(\mathrm{KH_2PO_4}\) and sulfuric acid \(\mathrm{H_2SO_4}\):\(\mathrm{KH_2PO_4(aq) + H_2SO_4(aq) \rightarrow KHSO_4(aq) + H_3PO_4(aq)}\) is balanced because the number of \(\mathrm{K}\), \(\mathrm{H}\), \(\mathrm{P}\), and \(\mathrm{O}\) atoms is the same on both sides.

To achieve balance:

• Count and list the number of atoms of each type for reactants and products.
• Adjust coefficients, which are the numbers in front of compounds, to equalize the atom counts.
• Iteratively adjust until complete balance is achieved.

ionic reactions

Ionic reactions involve ions, which are charged particles, and often occur when substances dissolve in water, disassociating into their ions. These reactions reveal the true nature of chemical interactions beyond molecular equations.

Take the reaction of barium hydroxide with sulfuric acid: \(\mathrm{Ba(OH)_2 + H_2SO_4 \rightarrow BaSO_4 + 4H_2O}\). When you rewrite this into ionic terms, you separate each compound into its ions. For instance,\(\mathrm{Ba^{2+} + 2OH^- + 2H^+ + SO_4^{2-} \rightarrow BaSO_4 + 4H_2O}\).

This representation helps us see which ions participate in the formation of the product and which ions are spectators. Understanding ionic reactions is crucial for predicting how ions will interact in different chemical environments, making it easier to manage reactions in a range of scientific fields. Remember:

• Ions form when compounds dissolve in water.
• Reactions often lead to the formation of precipitates or soluble products.
• Identifying participating ions allows for the simplification of reaction equations.