Question

1. a 3.69 g sample of sodium sulfate, na₂so₄, is dissolved in enough water to give 1.50 l of solution. calculate the sodium ion and sulfate ion concentrations in this solution. (molar mass of na₂so₄ = 142.0 g/mol)

a. na⁺ = 1.73 × 10⁻² m, so₄²⁻ = 1.73 × 10⁻² m
b. na⁺ = 3.46 × 10⁻² m, so₄²⁻ = 1.73 × 10⁻² m
c. na⁺ = 3.46 × 10⁻² m, so₄²⁻ = 3.46 × 10⁻² m
d. na⁺ = 1.73 × 10⁻² m, so₄²⁻ = 3.46 × 10⁻² m
e. na⁺ = 1.73 × 10⁻² m, so₄²⁻ = 8.66 × 10⁻³ m

Explanation:

Step1: Calculate the moles of \(Na_{2}SO_{4}\)

Use the formula \(n=\frac{m}{M}\), where \(m = 3.69g\) and \(M=142.0g/mol\).
\(n_{Na_{2}SO_{4}}=\frac{3.69g}{142.0g/mol}=2.60\times10^{- 2}mol\)

Step2: Calculate the molarity of \(Na_{2}SO_{4}\)

Use the formula \(M=\frac{n}{V}\), where \(n = 2.60\times10^{-2}mol\) and \(V = 1.50L\).
\(M_{Na_{2}SO_{4}}=\frac{2.60\times10^{-2}mol}{1.50L}=1.73\times10^{-2}M\)

Step3: Determine the ion - molarity relationships

From the dissociation equation \(Na_{2}SO_{4}(aq)
ightarrow2Na^{+}(aq)+SO_{4}^{2 - }(aq)\), we know that \([Na^{+}]=2[Na_{2}SO_{4}]\) and \([SO_{4}^{2 - }]=[Na_{2}SO_{4}]\)

\([Na^{+}]=2\times1.73\times10^{-2}M = 3.46\times10^{-2}M\)

\([SO_{4}^{2 - }]=1.73\times10^{-2}M\)

Answer:

B. \([Na^{+}]=3.46\times10^{-2}M\), \([SO_{4}^{2 - }]=1.73\times10^{-2}M\)