What is the molar concentration of sulfate ions in a 0.150 M $N a_{2} S O_{4}$ $Na_2SO_4$ solution?

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What is the molar concentration of sulfate ions in a 0.150 M $N a_{2} S O_{4}$ $Na_2SO_4$ solution?

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Answer 1 · 2016-05-28T15:43:33

$M_{S O_{4}^{2 -}} = 0.150 M$ $M_(SO_4^(2-))=0.150M$

Explanation:

In solution, sodium sulfate dissociates according to the following equation:

$N a_{2} S O_{4} (s) \to 2 N a^{+} (a q) + S O_{4}^{2 -} (a q)$ $Na_2SO_4(s)->2Na^(+)(aq)+SO_4^(2-)(aq)$

Therefore, $1 m o l$ $1mol$ of $N a_{2} S O_{4}$ $Na_2SO_4$ will give $1 m o l$ $1mol$ of $S O_{4}^{2 -}$ $SO_4^(2-)$ and thus, the molarity of sulfate ions will be equal to that of the sodium sulfate.

$M_{S O_{4}^{2 -}} = 0.150 M$ $M_(SO_4^(2-))=0.150M$

However, the molarity of sodium ions is the double of that of sodium sulfate since $1 m o l$ $1mol$ of $N a_{2} S O_{4}$ $Na_2SO_4$ will give $2 m o l$ $2mol$ of $N a^{+}$ $Na^(+)$ :

$M_{N a^{+}}^{+} = 0.300 M$ $M_(Na^(+))=0.300M$

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What is the molar concentration of sulfate ions in a 0.150 M $N a_{2} S O_{4}$ $Na_2SO_4$ solution?

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What is the molar concentration of sulfate ions in a 0.150 M Na_2SO_4Na2SO4Na_2SO_4 solution?

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What is the molar concentration of sulfate ions in a 0.150 M $N a_{2} S O_{4}$ $Na_2SO_4$ solution?