Question

suppose 28.7 g of potassium citrate (k₃c₆h₅o₇) are dissolved in 530. ml 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: Find the molar mass of \(K_3C_6H_5O_7\)

The molar mass of \(K\) is \(39.10\space g/mol\), \(C\) is \(12.01\space g/mol\), \(H\) is \(1.008\space g/mol\), and \(O\) is \(16.00\space g/mol\).

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Step2: Calculate the moles of \(K_3C_6H_5O_7\)

Use the formula \(n=\frac{m}{M}\), where \(m = 28.7\space g\) and \(M = 306.4\space g/mol\)

$$n=\frac{28.7\space g}{306.4\space g/mol}\approx0.0937\space mol$$

Step3: Write the dissociation equation

\(K_3C_6H_5O_7(s)\to3K^{+}(aq)+C_6H_5O_7^{3 -}(aq)\)

Step4: Calculate the moles of each ion

For \(K^{+}\): \(n_{K^{+}}=3\times0.0937\space mol = 0.281\space mol\)
For \(C_6H_5O_7^{3 -}\): \(n_{C_6H_5O_7^{3 -}}=0.0937\space mol\)

Step5: Convert volume to liters

\(V = 530\space mL=0.530\space L\)

Step6: Calculate the equivalent of each ion

For \(K^{+}\): Each \(K^{+}\) has a charge of \(+ 1\), so \(Eq_{K^{+}}=n_{K^{+}}\times1 = 0.281\space Eq\)

$$C_{K^{+}}=\frac{0.281\space Eq}{0.530\space L}\times1000\space mEq/Eq\approx530\space mEq/L$$

For \(C_6H_5O_7^{3 -}\): Each \(C_6H_5O_7^{3 -}\) has a charge of \(-3\), so \(Eq_{C_6H_5O_7^{3 -}}=n_{C_6H_5O_7^{3 -}}\times3=0.0937\times3 = 0.281\space Eq\)

$$C_{C_6H_5O_7^{3 -}}=\frac{0.281\space Eq}{0.530\space L}\times1000\space mEq/Eq\approx530\space mEq/L$$

Answer:

ion	\(\frac{mEq}{L}\)
\(C_6H_5O_7^{3 -}\)	\(530\)