Question

suppose 252 g of potassium phosphate (k₃po₄) are dissolved in 7.70 l of water. find the composition of the resulting electrolyte solution. in particular, list the chemical symbols (including any charge) of each dissolved ion in the table below. list only one ion per row. then, calculate the concentration of each ion in \\( \frac{meq}{l} \\), and write the concentration in the second column of each row. be sure you round your answers to the correct number of significant digits.

Explanation:

Step1: Calculate the molar mass of \(K_3PO_4\)

The molar mass of \(K\) is \(39.10\space g/mol\), of \(P\) is \(30.97\space g/mol\), and of \(O\) is \(16.00\space g/mol\).

$$M = 3\times39.10+30.97 + 4\times16.00=212.27\space g/mol$$

Step2: Calculate the number of moles of \(K_3PO_4\)

$$n=\frac{m}{M}=\frac{252\space g}{212.27\space g/mol}\approx1.187\space mol$$

Step3: Determine the dissociation of \(K_3PO_4\)

$$K_3PO_4(s) ightarrow3K^{+}(aq)+PO_4^{3 -}(aq)$$

Step4: Calculate the number of equivalents for each ion

For \(K^{+}\): Each \(K_3PO_4\) gives \(3\) \(K^{+}\) ions. The valence of \(K^{+}\) is \(1\).
The number of moles of \(K^{+}\) is \(n_{K^{+}}=3\times1.187 = 3.561\space mol\)
The number of equivalents of \(K^{+}\), \(Eq_{K^{+}}=3.561\space mol\times1 = 3561\space mEq\)
Concentration of \(K^{+}\), \([K^{+}]=\frac{3561\space mEq}{7.70\space L}\approx462\space mEq/L\)

For \(PO_4^{3-}\): Each \(K_3PO_4\) gives \(1\) \(PO_4^{3-}\) ion. The valence of \(PO_4^{3-}\) is \(3\).
The number of moles of \(PO_4^{3-}\) is \(n_{PO_4^{3-}}=1.187\space mol\)
The number of equivalents of \(PO_4^{3-}\), \(Eq_{PO_4^{3-}}=1.187\space mol\times3=3561\space mEq\)
Concentration of \(PO_4^{3-}\), \([PO_4^{3-}]=\frac{3561\space mEq}{7.70\space L}\approx462\space mEq/L\)

Answer:

ion	\(\frac{mEq}{L}\)
\(PO_4^{3-}\)	\(462\)