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Chapter 4: Problem 124

Complete and balance each acid-base reaction. a. \(\mathrm{H}_{3} \mathrm{PO}_{4}(a q)+\mathrm{NaOH}(a q) \rightarrow\) Contains three acidic hydrogens b. \(\mathrm{H}_{2} \mathrm{SO}_{4}(a q)+\mathrm{Al}(\mathrm{OH})_{3}(s) \rightarrow\) Contains two acidic hydrogens c. \(\mathrm{H}_{2} \mathrm{Se}(a q)+\mathrm{Ba}(\mathrm{OH})_{2}(a q) \rightarrow\) Contains two acidic hydrogens d. \(\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}(a q)+\mathrm{NaOH}(a q) \rightarrow\) Contains two acidic hydrogens.

Short Answer

Expert verified
The short version of the answers for each reaction is as follows: a. \(\mathrm{H}_{3} \mathrm{PO}_{4}(a q)+\mathrm{NaOH}(a q) \rightarrow \mathrm{NaH}_{2}\mathrm{PO}_{4}(a q) + \mathrm{H}_{2}\mathrm{O}(l)\) b. \(2 \mathrm{H}_{2} \mathrm{SO}_{4}(a q)+\mathrm{Al}_{2}(\mathrm{OH})_{6}(s) \rightarrow \mathrm{Al}_{2}(\mathrm{SO}_{4})_{3}(a q) + 6 \mathrm{H}_{2}\mathrm{O}(l)\) c. \(\mathrm{H}_{2} \mathrm{Se}(a q)+\mathrm{Ba}(\mathrm{OH})_{2}(a q) \rightarrow \mathrm{BaSe}(s) + \mathrm{H}_{2}\mathrm{O}(l)\) d. \(\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}(a q)+2\mathrm{NaOH}(a q) \rightarrow \mathrm{Na}_{2}\mathrm{C}_{2}\mathrm{O}_{4}(a q) + 2\mathrm{H}_{2}\mathrm{O}(l)\)

Step by step solution

01

Identify the products

Exchange one acidic hydrogen ion (H+) from H₃PO₄ and a hydroxide ion (OH-) from NaOH to form water (H₂O) and the other product, sodium phosphate (NaH₂PO₄). So the reaction is: \[\mathrm{H}_{3} \mathrm{PO}_{4}(a q)+\mathrm{NaOH}(a q) \rightarrow \mathrm{NaH}_{2}\mathrm{PO}_{4}(a q) + \mathrm{H}_{2}\mathrm{O}(l)\]
02

Balance the equation

The current equation has equal atoms of each element on both sides, so it is already balanced. The final balanced equation is: \[\mathrm{H}_{3} \mathrm{PO}_{4}(a q)+\mathrm{NaOH}(a q) \rightarrow \mathrm{NaH}_{2}\mathrm{PO}_{4}(a q) + \mathrm{H}_{2}\mathrm{O}(l)\] b. \(\mathrm{H}_{2} \mathrm{SO}_{4}(a q)+\mathrm{Al}(\mathrm{OH})_{3}(s) \rightarrow\)
03

Identify the products

Exchange two acidic hydrogen ions (H+) from H₂SO₄ and the hydroxide ions (OH-) from Al(OH)₃ to form water (H₂O) and the other product, aluminum sulfate (Al₂(SO₄)₃): So the reaction is: \[\mathrm{H}_{2} \mathrm{SO}_{4}(a q)+\mathrm{Al}(\mathrm{OH})_{3}(s) \rightarrow \mathrm{Al}_{2}(\mathrm{SO}_{4})_{3}(a q) + \mathrm{H}_{2}\mathrm{O}(l)\]
04

Balance the equation

We need to balance the equation by adjusting the coefficients of the reactants and products, so we get: \[2 \mathrm{H}_{2} \mathrm{SO}_{4}(a q)+\mathrm{Al}_{2}(\mathrm{OH})_{6}(s) \rightarrow \mathrm{Al}_{2}(\mathrm{SO}_{4})_{3}(a q) + 6 \mathrm{H}_{2}\mathrm{O}(l)\] c. \(\mathrm{H}_{2} \mathrm{Se}(a q)+\mathrm{Ba}(\mathrm{OH})_{2}(a q) \rightarrow\)
05

Identify the products

Exchange two acidic hydrogen ions (H+) from H₂Se and the hydroxide ions (OH-) from Ba(OH)₂ to form water (H₂O) and the other product, barium selenide (BaSe): So the reaction is: \[\mathrm{H}_{2} \mathrm{Se}(a q)+\mathrm{Ba}(\mathrm{OH})_{2}(a q) \rightarrow \mathrm{BaSe}(s) + \mathrm{H}_{2}\mathrm{O}(l)\]
06

Balance the equation

The current equation has equal atoms of each element on both sides, so it is already balanced. The final balanced equation is: \[\mathrm{H}_{2} \mathrm{Se}(a q)+\mathrm{Ba}(\mathrm{OH})_{2}(a q) \rightarrow \mathrm{BaSe}(s) + \mathrm{H}_{2}\mathrm{O}(l)\] d. \(\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}(a q)+\mathrm{NaOH}(a q) \rightarrow\)
07

Identify the products

Exchange two acidic hydrogen ions (H+) from H₂C₂O₄ and the hydroxide ions (OH-) from NaOH to form water (H₂O) and the other product, sodium oxalate (Na₂C₂O₄): So the reaction is: \[\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}(a q)+\mathrm{NaOH}(a q) \rightarrow \mathrm{Na}_{2}\mathrm{C}_{2}\mathrm{O}_{4}(a q) + \mathrm{H}_{2}\mathrm{O}(l)\]
08

Balance the equation

We need to balance the equation by adjusting the coefficients of the reactants and products, so we get: \[\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}(a q)+2\mathrm{NaOH}(a q) \rightarrow \mathrm{Na}_{2}\mathrm{C}_{2}\mathrm{O}_{4}(a q) + 2\mathrm{H}_{2}\mathrm{O}(l)\]

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Most popular questions from this chapter

A student mixes four reagents together, thinking that the solutions will neutralize each other. The solutions mixed together are \(50.0 \mathrm{~mL}\) of \(0.100 M\) hydrochloric acid, \(100.0 \mathrm{~mL}\) of \(0.200 M\) of nitric acid, \(500.0 \mathrm{~mL}\) of \(0.0100 \mathrm{M}\) calcium hydroxide, and \(200.0 \mathrm{~mL}\) of \(0.100 M\) rubidium hydroxide. Did the acids and bases exactly neutralize each other? If not, calculate the concentration of excess \(\mathrm{H}^{+}\) or \(\mathrm{OH}^{-}\) ions left in solution.

What mass of solid aluminum hydroxide can be produced when \(50.0 \mathrm{~mL}\) of \(0.200 \mathrm{M} \mathrm{Al}\left(\mathrm{NO}_{3}\right)_{3}\) is added to \(200.0 \mathrm{~mL}\) of \(0.100 \mathrm{M} \mathrm{KOH} ?\)

A solution was prepared by mixing \(50.00 \mathrm{~mL}\) of \(0.100 \mathrm{M} \mathrm{HNO}_{3}\) and \(100.00 \mathrm{~mL}\) of \(0.200 \mathrm{M} \mathrm{HNO}_{3} .\) Calculate the molarity of the final solution of nitric acid.

Consider the reaction of \(19.0 \mathrm{~g}\) of zinc with excess silver nitrite to produce silver metal and zinc nitrite. The reaction is stopped before all the zinc metal has reacted and \(29.0 \mathrm{~g}\) of solid metal is present. Calculate the mass of each metal in the \(29.0-\mathrm{g}\) mixture.

Assign oxidation states for all atoms in each of the following compounds. a. \(\mathrm{KMnO}_{4}\) f. \(\mathrm{Fe}_{2} \mathrm{O}_{4}\) \(\begin{array}{ll}\text { b. } \mathrm{NiO}_{2} & \text { g. } \mathrm{XeOF}_{4}\end{array}\) c. \(\mathrm{Na}_{4} \mathrm{Fe}(\mathrm{OH})_{6}\) h. \(\mathrm{SF}_{4}\) d. \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{HPO}_{4}\) i. \(\mathrm{CO}\) e. \(\mathrm{P}_{4} \mathrm{O}_{6}\) j. \(\mathrm{C}_{6} \mathrm{H}_{1}, \mathrm{O}_{e}\)
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