Question

Complete and balance each acid-base reaction. a. ${\mathrm{H}}_3{\mathrm{PO}}_4(aq)+\mathrm{NaOH}(aq)\to$ Contains three acidic hydrogens b. ${\mathrm{H}}_2{\mathrm{SO}}_4(aq)+\mathrm{Al}(\mathrm{OH}{)}_3(s)\to$ Contains two acidic hydrogens c. ${\mathrm{H}}_2\mathrm{Se}(aq)+\mathrm{Ba}(\mathrm{OH}{)}_2(aq)\to$ Contains two acidic hydrogens d. ${\mathrm{H}}_2{\mathrm{C}}_2{\mathrm{O}}_4(aq)+\mathrm{NaOH}(aq)\to$

Short answer

a. $H_{3}P{O}_4(aq)+3NaOH(aq)\to N{a}_{3}P{O}_4(aq)+3H_{2}O(l)$ b. $3H_{2}S{O}_4(aq)+2Al(OH{)}_3(s)\to A{l}_2(S{O}_4{)}_3(aq)+6H_{2}O(l)$ c. $H_{2}Se(aq)+Ba(OH{)}_2(aq)\to BaSe(aq)+2H_{2}O(l)$ d. $H_2{C}_2{O}_4(aq)+2NaOH(aq)\to N{a}_2{C}_2{O}_4(aq)+2H_{2}O(l)$

Step-by-step solution

a. Complete and balance the H3PO4 and NaOH reaction:

To complete this reaction, we need to identify the products. When an acid reacts with a base, the resulting products are a salt and water. In this case, phosphoric acid (H3PO4) reacts with sodium hydroxide (NaOH). Their reaction will produce sodium phosphate (Na3PO4) and water (H2O). Next, we must balance the equation. The balanced equation for this reaction is: $$H_{3}P{O}_4(aq)+3NaOH(aq)\to N{a}_{3}P{O}_4(aq)+3H_{2}O(l)$$

b. Complete and balance the H₂SO₄ and Al(OH)₃ reaction:

This reaction will produce aluminum sulfate (Al₂(SO₄)₃) and water (H2O). The balanced equation for this reaction is: $$3H_{2}S{O}_4(aq)+2Al(OH{)}_3(s)\to A{l}_2(S{O}_4{)}_3(aq)+6H_{2}O(l)$$

c. Complete and balance the H₂Se and Ba(OH)₂ reaction:

This reaction will produce barium selenide (BaSe) and water (H2O). The balanced equation for this reaction is: $$H_{2}Se(aq)+Ba(OH{)}_2(aq)\to BaSe(aq)+2H_{2}O(l)$$

d. Complete and balance the H₂C₂O₄ and NaOH reaction:

This reaction will produce sodium oxalate (Na₂C₂O₄) and water (H2O). The balanced equation for this reaction is: $$H_2{C}_2{O}_4(aq)+2NaOH(aq)\to N{a}_2{C}_2{O}_4(aq)+2H_{2}O(l)$$

Key concepts

Chemical Equation Balancing

Balancing chemical equations is similar to the art of ensuring both sides of a scale weigh the same. In a balanced chemical equation, the number of atoms for each element must be equal on the reactant and product sides. This maintains the law of conservation of mass, which states that mass is neither created nor destroyed in a chemical reaction.

For example, in the reaction of phosphoric acid (H₃PO₄) with sodium hydroxide (NaOH), we balance the equation by ensuring that there are equal numbers of hydrogen, oxygen, phosphorus, and sodium atoms on both sides. We find that we need three NaOH molecules to balance the three hydrogen atoms in H₃PO₄, resulting in the balanced reaction: $$H_{3}P{O}_4(aq)+3NaOH(aq)\to N{a}_{3}P{O}_4(aq)+3H_{2}O(l)$$

Tips for Balancing Equations

• Start by balancing the atoms of elements that appear only once on each side.
• Balance hydrogen and oxygen atoms last, as they often appear in multiple compounds.
• Adjust coefficients, numbers in front of compounds, to balance atoms rather than changing subscripts, which can alter the compound's identity.

Acid-Base Neutralization

Acid-base neutralization is the heart of a titration. Imagine it as a dance between acids and bases where they come together to form water and salt. An acid carries hydrogen ions (H⁺) while a base come equipped with hydroxide ions (OH^-). When they react, they form water (H₂O) and an ionic compound known as a salt.

For instance, the neutralization reaction between sulfuric acid (H₂SO₄) and aluminum hydroxide [Al(OH)₃] leads to:$$3H_{2}S{O}_4(aq)+2Al(OH{)}_3(s)\to A{l}_2(S{O}_4{)}_3(aq)+6H_{2}O(l)$$Here, we witness each pair of acid and base yielding their ions, which marry to produce the salt aluminum sulfate (Al₂(SO₄)₃) and water.

Neutralization reactions are pivotal in various applications such as antacid tablets combating stomach acid, or environmental strategies to neutralize acidic lake water.

Stoichiometry

Stoichiometry might sound complex, but it's simply the calculation of reactants and products in chemical reactions. It is the recipe for chemistry, providing precise measurements to get just the right outcome. This mathematical method of stoichiometry uses the balanced equation as a basis to compare amounts in moles.

Consider the stoichiometric relationship in the reaction of hydrogen selenide (H₂Se) and barium hydroxide (Ba(OH)₂), which is described by the equation:$$H_{2}Se(aq)+Ba(OH{)}_2(aq)\to BaSe(aq)+2H_{2}O(l)$$The coefficients in the balanced equation tell us that one mole of H₂Se reacts with one mole of Ba(OH)₂ to produce one mole of BaSe and two moles of water. Stoichiometry helps us predict the amount of each substance needed or formed in a reaction, vital for applications in chemical manufacturing and understanding natural processes.

Salt Formation in Reactions

When we hear 'salt', we often think of the common table salt, but in chemistry, salts are so much more. They are the offspring of acid-base neutralization reactions. A salt is an ionic compound composed of a cation from a base and an anion from an acid.

For example, sodium oxalate (Na₂C₂O₄) is a salt that forms when oxalic acid (H₂C₂O₄) reacts with sodium hydroxide (NaOH), as shown:$$H_2{C}_2{O}_4(aq)+2NaOH(aq)\to N{a}_2{C}_2{O}_4(aq)+2H_{2}O(l)$$The sodium ions (Na⁺) from NaOH replace the hydrogen ions (H⁺) from oxalic acid, and voilà, we get sodium oxalate. Such salt formation is fundamental in countless products, from soaps and detergents to medications and fertilizers, illustrating the immense practicality of understanding these reactions.