Calculate the number of equivalents per liter (eq/L) of phosphate ion, [latex]PO_4 ^{3–}[/latex], in a solution that is [latex]5.0 \times 10^{–3}[/latex] M phosphate.

Step 1. It is necessary to use two conversion factors: mol [latex]PO_4 ^{3–}[/latex] → mol charge and mol charge → eq [latex]PO_4 ^{3–}[/latex] Step 2. Arranging these factors in sequence yields: [latex]\frac{5.0 \times 10^{-3} \cancel{mol PO_4 ^{3–}}}{1 L} \times \frac{3 \cancel{mol charge}}{1 \cancel{mol PO_4 ^{3–}}} \times \frac{1 eq PO_4 ^{3–}}{1 \cancel{mol charge}} = \frac{1.5 \times 10^{-2} eq PO_4 ^{3–}}{L}[/latex]

Question 6.10: Calculate the number of equivalents per liter (eq/L) of phos...

General, Organic and Biochemistry

Calculate the number of equivalents per liter (eq/L) of phosphate ion, $PO_4 ^{3–}$ , in a solution that is $5.0 \times 10^{–3}$ M phosphate.

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Step 1. It is necessary to use two conversion factors:
mol $PO_4 ^{3–}$ → mol charge
and
mol charge → eq $PO_4 ^{3–}$

Step 2. Arranging these factors in sequence yields:
$\frac{5.0 \times 10^{-3} \cancel{mol PO_4 ^{3–}}}{1 L} \times \frac{3 \cancel{mol charge}}{1 \cancel{mol PO_4 ^{3–}}} \times \frac{1 eq PO_4 ^{3–}}{1 \cancel{mol charge}} = \frac{1.5 \times 10^{-2} eq PO_4 ^{3–}}{L}$