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Chapter 15: Problem 132

A sodium dihydrogen phosphate solution was prepared by dissolving \(5.0 \mathrm{~g}\) of \(\mathrm{NaH}_{2} \mathrm{PO}_{4}\) in enough water to make \(500 . \mathrm{mL}\) of solution. What are the molarity and normality of the resulting solution?

Short Answer

Expert verified
The molarity of the sodium dihydrogen phosphate solution is 0.0834 M, and the normality is 0.0834 N.

Step by step solution

01

Calculate the Molar Mass of NaH鈧侾O鈧�

First, we need to find the molar mass of sodium dihydrogen phosphate, NaH鈧侾O鈧�. To do this, we must sum the molar masses of its constituent elements: - Sodium (Na): 22.99 g/mol - Hydrogen (H): 1.01 g/mol - Phosphorus (P): 30.97 g/mol - Oxygen (O): 16.00 g/mol So the molar mass of NaH鈧侾O鈧� is: \(1\times(22.99) + 2\times(1.01) + 1\times(30.97) + 4\times(16.00) = 119.98 \, \text{g/mol}\)
02

Calculate the Number of Moles of NaH鈧侾O鈧�

Now we can calculate the number of moles in the 5.0 g of NaH鈧侾O鈧�: \(\text{moles} = \frac{\text{mass}}{\text{molar mass}} = \frac{5.0}{119.98} = 0.0417 \, \text{mol}\)
03

Calculate the Molarity of the Solution

Now we can calculate the molarity of the solution. Molarity is defined as the number of moles of solute dissolved per liter of solution. First, we need to convert the volume of the solution from mL to L: \(\text{volume} = 500 \, \text{mL} \cdot \frac{1 \, \text{L}}{1000 \, \text{mL}} = 0.500 \, \text{L}\) Now, we can calculate the molarity: \(\text{Molarity} = \frac{\text{moles}}{\text{volume}} = \frac{0.0417}{0.500} = 0.0834 \, \text{M}\)
04

Calculate the Normality of the Solution

Normality is defined as the number of gram-equivalents of solute dissolved per liter of solution. In this case, NaH鈧侾O鈧� can donate 1 proton (H鈦�) in the solution, so the normality is equal to the molarity: \(\text{Normality} = 0.0834 \, \text{N}\) So, the molarity of the sodium dihydrogen phosphate solution is 0.0834 M, and the normality is 0.0834 N.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Molarity Calculation
Understanding the molarity of a solution is essential in many chemistry applications. Molarity (M) relates to concentration and represents the number of moles of solute per liter of solution.
To find molarity, you first identify the number of moles of the solute, which is sodium dihydrogen phosphate (\(\text{NaH}_2\text{PO}_4\)) in this exercise. This compound has a calculated molar mass of 119.98 g/mol.
After determining the moles from a given mass of the solute, you divide this quantity by the solution's total volume in liters.
In our example, we calculated that 5.0 grams of \(\text{NaH}_2\text{PO}_4\) results in 0.0417 moles. Since the volume of this solution is given as 500 mL (or 0.500 L), molarity becomes:
  • Moles of solute: 0.0417 mol
  • Volume of solution: 0.500 L
  • Molarity: \(\frac{0.0417}{0.500} = 0.0834 \, \text{M}\)
This results in a molarity of 0.0834 M for the solution, indicating the concentration of \(\text{NaH}_2\text{PO}_4\) in this solution.
Normality Calculation
Normality (N) is another way to express the concentration of a solution and can often be more directly related to the chemical reactions involving acids, bases, or redox reactions. It refers to the number of gram-equivalents of solute per liter of solution.
For certain reactions, particularly acid-base reactions, normality directly relates to molarity. In the case of sodium dihydrogen phosphate (\(\text{NaH}_2\text{PO}_4\)), the substance can donate one proton (\(\text{H}^+\)) in a solution.
Thus, for \(\text{NaH}_2\text{PO}_4\), the normality equals the molarity because it acts as a monovalent solute (1 equivalent per mole).
Since the molarity of our \(\text{NaH}_2\text{PO}_4\) solution was calculated as 0.0834 M, the normality is the same:
  • Normality: 0.0834 N
This means each liter of solution contains 0.0834 gram-equivalents of hydrogen ions.
Sodium Dihydrogen Phosphate
Sodium dihydrogen phosphate (\(\text{NaH}_2\text{PO}_4\)) is an example of a compound that behaves as an acid when dissolved in water. Unlike neutral salts, it contributes \(\text{H}^+\) ions to the solution.
This characteristic makes it useful in various applications, especially in buffering solutions where maintaining a stable pH is critical.
This compound is part of the phosphate group and is commonly encountered in laboratory and industrial processes. Here are a few key things to know:
  • It has a known molar mass of 119.98 g/mol.
  • When dissolved in water, it primarily releases one \(\text{H}^+\), impacting \(\text{pH}\) and solution behavior.
  • Applications include its role in nutritional supplements, detergents, and as a leavening agent in baking.
Understanding its behavior, especially its impact on acidity (\(\text{H}^+\) release), is crucial for its effective use in chemical solutions.

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Most popular questions from this chapter

Calculate the new molarity when \(150 . \mathrm{mL}\) of water is added to each of the following solutions. a. \(125 \mathrm{~mL}\) of \(0.200 \mathrm{M} \mathrm{HBr}\) b. \(155 \mathrm{~mL}\) of \(0.250 \mathrm{M} \mathrm{Ca}\left(\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\right)_{2}\) c. \(0.500 \mathrm{~L}\) of \(0.250 \mathrm{M} \mathrm{H}_{3} \mathrm{PO}_{4}\) d. \(15 \mathrm{~mL}\) of \(18.0 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4}\)

If \(500 .\) g of water is added to \(75 \mathrm{~g}\) of \(25 \% \mathrm{NaCl}\) solution, what is the percent by mass of \(\mathrm{NaCl}\) in the diluted solution?

Calculate the number of moles of each ion present in each of the following solutions. a. \(1.25 \mathrm{~L}\) of \(0.250 \mathrm{M} \mathrm{Na}_{3} \mathrm{PO}_{4}\) solution b. \(3.5 \mathrm{~mL}\) of \(6.0 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4}\) solution c. \(25 \mathrm{~mL}\) of \(0.15 \mathrm{M} \mathrm{AlCl}_{3}\) solution d. 1.50 L of \(1.25 M\) BaCl \(_{2}\) solution

Calculate the new molarity if each of the following dilutions is made. Assume the volumes are additive. a. \(55.0 \mathrm{~mL}\) of water is added to \(25.0 \mathrm{~mL}\) of \(0.119 \mathrm{M} \mathrm{NaCl}\) solution b. \(125 \mathrm{~mL}\) of water is added to \(45.3 \mathrm{~mL}\) of \(0.701 \mathrm{M} \mathrm{NaOH}\) solution c. \(550 . \mathrm{mL}\) of water is added to \(125 \mathrm{~mL}\) of \(3.01 \mathrm{M}\) KOH solution d. \(335 \mathrm{~mL}\) of water is added to \(75.3 \mathrm{~mL}\) of \(2.07 \mathrm{M} \mathrm{CaCl}_{2}\) solution

Calculate the normality of each of the following solutions. a. \(0.134 M \mathrm{NaOH}\) b. \(0.00521 M \mathrm{Ca}(\mathrm{OH})_{2}\) c. \(4.42 M \mathrm{H}_{3} \mathrm{PO}_{4}\)
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