Question

how many liters of a 5.00 m solution of na₃po₄ are needed to have 4.00 equivalents ?
first we need the equation for dissolving the salt, na₃po₄ → 3 na⁺¹ + po₄⁻³
an equivalent is a mole of charges, either + or -, but not both, they are not added
the conversion factor is 1 mol = 1 eq
first convert equivalents into moles
1 mol
(2 eq)(---------) = 3 mol
4 eq
then calculate the liters
1 l
(5 mol)(-----------) = 6 l
7 mol
a. 0.00 b. 1.00 c. 2.00 d. 3.00 e. 4.00 f. 5.00
g. 6.00 h. 8.00 i. 10.00 j. 2000 k. 10² l. 10³
m. 10⁶ n. 10⁹ o. 10¹² p. 10⁻² q. 10⁻³ r. 10⁻⁶
s. 10⁻⁹ t. 10⁻¹² u. 10⁸ v. 10⁻¹ w. 10¹ x. 30.0
y. 0.100 z. 0.200 aa. 20.0 bb. 40.0
cc. 1.33 dd. 0.267

Explanation:

Step1: Determine equivalents to moles conversion

From the dissociation \( \ce{Na3PO4 -> 3Na+ + PO4^{3-}} \), the total charge per mole of \( \ce{Na3PO4} \) is \( 3(+1) + (-3) = 0 \), but for equivalents, we consider the charge of the ion. For \( \ce{PO4^{3-}} \), each mole of \( \ce{Na3PO4} \) provides 3 equivalents of negative charge (or 3 equivalents from \( \ce{Na+} \) as positive charge). Wait, the problem states the conversion factor is 1 mol = 1 eq? Wait, no, let's re - examine. Wait, the problem says "the conversion factor is 1 mol = __1__ eq" – maybe a typo, but looking at the dissociation, each mole of \( \ce{Na3PO4} \) gives 3 moles of \( \ce{Na+} \) (each \( \ce{Na+} \) is 1 equivalent) and 1 mole of \( \ce{PO4^{3-}} \) (3 equivalents). But the problem's first step: convert 4.00 equivalents to moles. Wait, the problem's first blank (2) is the number of equivalents, which is 4.00 (E option). Then the conversion factor: if we consider that for \( \ce{Na3PO4} \), the number of equivalents per mole is 3 (from \( \ce{Na+} \) or \( \ce{PO4^{3-}} \)). Wait, maybe the problem has a different approach. Wait, the first step: \( (4.00\ \text{eq}) \times \frac{1\ \text{mol}}{3\ \text{eq}} \)? No, the problem's given conversion factor is 1 mol = 1 eq? Wait, the problem's first calculation: \( (\_2\_\ \text{eq}) \times \frac{1\ \text{mol}}{\_4\_\ \text{eq}} = \_3\_\ \text{mol} \). So \( \_2\_ = 4.00 \) (E), \( \_4\_ = 3 \)? Wait, no, maybe the problem is considering that each mole of \( \ce{Na3PO4} \) has 3 equivalents (from \( \ce{Na+} \) or \( \ce{PO4^{3-}} \)). Wait, let's do it step by step.

First, find moles of \( \ce{Na3PO4} \) from equivalents. The dissociation is \( \ce{Na3PO4 -> 3Na+ + PO4^{3-}} \). The number of equivalents per mole of \( \ce{Na3PO4} \) is 3 (because each \( \ce{Na+} \) is 1 equivalent, 3 \( \ce{Na+} \) per mole, or \( \ce{PO4^{3-}} \) is 3 equivalents per mole). So the conversion factor is \( \frac{1\ \text{mol}}{3\ \text{eq}} \). But the problem's first step has \( \frac{1\ \text{mol}}{\_4\_\ \text{eq}} \), so \( \_4\_ = 3 \)? But the options don't have 3 as a denominator? Wait, maybe the problem is simplified. Wait, the problem's first calculation: \( (4.00\ \text{eq}) \times \frac{1\ \text{mol}}{3\ \text{eq}}=\frac{4}{3}\approx1.33\ \text{mol} \)? No, wait, the problem's next step is calculating liters: \( (\_5\_\ \text{mol}) \times \frac{1\ \text{L}}{5.00\ \text{mol}} \). So \( \_5\_ \) is the moles from the first step. Let's re - express:

Molarity \( M=\frac{\text{moles of solute}}{\text{volume of solution (L)}} \), so \( V = \frac{n}{M} \).

First, find moles \( n \) from equivalents. For \( \ce{Na3PO4} \), the number of equivalents per mole (\( \text{equiv/mol} \)) is equal to the total charge of the cation or anion. \( \ce{Na+} \) has charge +1, 3 per mole, so 3 equiv/mol. \( \ce{PO4^{3-}} \) has charge - 3, 1 per mole, so 3 equiv/mol. So \( \text{moles}=\frac{\text{equivalents}}{\text{equiv/mol}}=\frac{4.00\ \text{eq}}{3\ \text{eq/mol}}\approx1.33\ \text{mol} \)? No, wait, the problem's first step: the blank (2) is 4.00 (E), blank (4) is 3? But the options for blank (4) – no, the options are given as A - DD. Wait, maybe the problem has a mistake, but let's follow the problem's structure.

Wait, the problem's first step: \( (\_2\_\ \text{eq}) \times \frac{1\ \text{mol}}{\_4\_\ \text{eq}}=\_3\_\ \text{mol} \). So \( \_2\_ = 4.00 \) (E), \( \_4\_ = 3 \)? But 3 is not an option. Wait, maybe the problem considers that 1 mol = 1 eq, which is wrong, but let's proceed with the problem's given. If 1 mol = 1 eq, then \( 4.00\ \text{eq} \times \frac{1\ \text{mol}}{1\ \text{eq}} = 4.00\ \text{mol} \), but then the next step: \( (\_5\_\ \text{mol}) \times \frac{1\ \text{L}}{5.00\ \text{mol}} \). So \( \_5\_ = 4.00 \), then \( 4.00\ \text{mol} \times \frac{1\ \text{L}}{5.00\ \text{mol}} = 0.800\ \text{L} \), which is not an option. So my initial approach is wrong.

Wait, let's re - read the problem: "An equivalent is a mole of charges, either + or -, but not both, they are not added the conversion factor is 1 mol = __1__ eq". Wait, maybe for \( \ce{Na3PO4} \), when it dissociates into \( 3\ce{Na+} \) and \( \ce{PO4^{3-}} \), the number of equivalents per mole is 3 (from \( \ce{Na+} \): 3 moles of \( \ce{Na+} \), each is 1 equivalent, so 3 equivalents; or from \( \ce{PO4^{3-}} \): 1 mole of \( \ce{PO4^{3-}} \), 3 equivalents). So the conversion factor is \( \text{mol}=\frac{\text{equivalents}}{3} \). So \( \text{moles}=\frac{4.00\ \text{eq}}{3\ \text{eq/mol}}\approx1.33\ \text{mol} \) (CC option). Then, to find volume: \( V=\frac{n}{M}=\frac{1.33\ \text{mol}}{5.00\ \text{mol/L}}\approx0.267\ \text{L} \) (DD option). Wait, let's check:

First step: convert equivalents to moles. \( 4.00\ \text{eq} \times \frac{1\ \text{mol}}{3\ \text{eq}}=\frac{4}{3}\approx1.33\ \text{mol} \) (so \( \_2\_ = 4.00 \) (E), \( \_4\_ = 3 \) (but 3 is not an option, maybe the problem has a typo, and the conversion factor is 1 mol = 3 eq? No, the problem says "1 mol = __1__ eq" – maybe the problem is considering that the equivalent weight is based on the cation or anion with charge 1, but no. Alternatively, maybe the problem is for a monovalent ion, but \( \ce{Na3PO4} \) is not.

Wait, let's do it correctly. The formula for equivalents: \( \text{Equivalents}=\text{Moles}\times\text{Number of charges} \). For \( \ce{Na3PO4} \), when it dissociates, \( \ce{Na3PO4 -> 3Na+ + PO4^{3-}} \). The number of equivalents per mole of \( \ce{Na3PO4} \) is 3 (from \( \ce{Na+} \)) or 3 (from \( \ce{PO4^{3-}} \)). So \( \text{Moles}=\frac{\text{Equivalents}}{\text{Number of charges}} \). So \( \text{Moles}=\frac{4.00\ \text{eq}}{3\ \text{eq/mol}}\approx1.33\ \text{mol} \). Then, volume \( V = \frac{n}{M}=\frac{1.33\ \text{mol}}{5.00\ \text{mol/L}}\approx0.267\ \text{L} \).

Now, let's fill the blanks:

Blank (2): 4.00 (E)

Blank (4): 3 (but 3 is not an option, maybe the problem considers 1 mol = 1 eq, which is wrong, but if we take 1 mol = 1 eq, then:

Blank (2): 4.00 (E)

Blank (4): 1 (but 1 is not an option? Wait, the options for the denominators: the options are A - DD. Wait, blank (4) is the denominator in the first conversion, so the options for blank (4) – looking at the options, 3 is not there, but maybe the problem has a mistake and the conversion factor is 1 mol = 3 eq, so \( \frac{1\ \text{mol}}{3\ \text{eq}} \), and 3 is not an option, but let's proceed with the correct calculation.

First, moles from equivalents: \( n=\frac{4.00\ \text{eq}}{3\ \text{eq/mol}}\approx1.33\ \text{mol} \) (CC)

Then, volume: \( V = \frac{n}{M}=\frac{1.33\ \text{mol}}{5.00\ \text{mol/L}}\approx0.267\ \text{L} \) (DD)

Now, let's check the problem's steps:

First step: \( (\_2\_\ \text{eq}) \times \frac{1\ \text{mol}}{\_4\_\ \text{eq}}=\_3\_\ \text{mol} \)

\( \_2\_ = 4.00 \) (E)

\( \_4\_ = 3 \) (not an option, but maybe the problem meant 3 eq per mol, so \( \_4\_ = 3 \) (but 3 is not in the options, maybe a mistake, but let's assume that the problem has \( \_4\_ = 3 \) (even though not an option), then \( \_3\_=\frac{4.00}{3}\approx1.33 \) (CC)

Second step: \( (\_5\_\ \text{mol}) \times \frac{1\ \text{L}}{\_7\_\ \text{mol}}=\_6\_\ \text{L} \)

\( \_5\_ = 1.33 \) (CC)

\( \_7\_ = 5.00 \) (F)

\( \_6\_=\frac{1.33}{5.00}\approx0.267 \) (DD)

Step2: Calculate liters from moles

Using the molarity formula \( M=\frac{n}{V} \), so \( V = \frac{n}{M} \). We found \( n\approx1.33\ \text{mol} \) and \( M = 5.00\ \text{mol/L} \), so \( V=\frac{1.33\ \text{mol}}{5.00\ \text{mol/L}} = 0.267\ \text{L} \)

Answer:

DD. 0.267