Question

suppose 60.4 g of sodium acetate (nach₃co₂) are dissolved in 6.40 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: Determine the ions formed

Sodium acetate \((\text{NaCH}_3\text{CO}_2)\) dissociates in water as \(\text{NaCH}_3\text{CO}_2
ightarrow\text{Na}^++\text{CH}_3\text{CO}_2^-\).

Step2: Calculate the molar mass of \(\text{NaCH}_3\text{CO}_2\)

The molar mass of \(\text{NaCH}_3\text{CO}_2\):

• \(M(\text{Na}) = 22.99\space g/mol\), \(M(\text{C})=12.01\space g/mol\), \(M(\text{H}) = 1.008\space g/mol\), \(M(\text{O})=16.00\space g/mol\)
• \(M(\text{NaCH}_3\text{CO}_2)=22.99+(2\times12.01)+(3\times1.008)+(2\times16.00)=82.034\space g/mol\)

Step3: Calculate the number of moles of \(\text{NaCH}_3\text{CO}_2\)

Using the formula \(n=\frac{m}{M}\), where \(m = 60.4\space g\) and \(M = 82.034\space g/mol\)

• \(n=\frac{60.4}{82.034}\approx0.736\space mol\)

Step4: Calculate the equivalent weight of each ion

• For \(\text{Na}^+\): The charge \(z = 1\), equivalent weight \(E=\frac{M(\text{Na})}{z}=\frac{22.99}{1}=22.99\space g/eq\)
• For \(\text{CH}_3\text{CO}_2^-\): The charge \(z = 1\), equivalent weight \(E=\frac{M(\text{CH}_3\text{CO}_2)}{z}=\frac{(2\times12.01)+(3\times1.008)+(2\times16.00)}{1}=59.044\space g/eq\)

Step5: Calculate the number of equivalents of \(\text{NaCH}_3\text{CO}_2\)

Since \(n = 0.736\space mol\) and for each mole of \(\text{NaCH}_3\text{CO}_2\) there is 1 equivalent of \(\text{Na}^+\) and 1 equivalent of \(\text{CH}_3\text{CO}_2^-\) (because \(z = 1\) for both ions)

• Number of equivalents of \(\text{Na}^+\) and \(\text{CH}_3\text{CO}_2^-\) is \(n_{eq}=0.736\space eq\)

Step6: Calculate the concentration in \(mEq/L\)

Using the formula \(C=\frac{n_{eq}\times1000}{V}\), where \(V = 6.40\space L\)

• For \(\text{Na}^+\): \(C=\frac{0.736\times1000}{6.40}\approx115\space mEq/L\)
• For \(\text{CH}_3\text{CO}_2^-\): \(C=\frac{0.736\times1000}{6.40}\approx115\space mEq/L\)

Answer:

ion	\(\frac{mEq}{L}\)
\(\text{CH}_3\text{CO}_2^-\)	\(115\)