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If 50.0 mL of 0.0160 M mercury(II) nitrate is combined with 23.0 mL of 0.120 M sodium sulfide, what is the resulting concentration of sodium ions in the solution after the reaction is complete? (Assume the solution volumes are additive). The balanced equation for the reaction is:

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sebassandin

Answer:

0.076M

Explanation:

Hello,

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Just some clarifications:

- the moles of sodium cations qre computed from the moles of sodium sulfide, taking into account that 2 mol of sodium atoms are present per mole of sodium sulfide.

- the volumes of the solutions must be added to get the total one.

Best regards.

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