Question #21606

Mercury compounds are poisonous, but mercury ions can be removed from a solution by precipitating insoluble mercury (II) sulfide, HgS (s). Determine the minimum volume of 0.0783 mol/L sodium sulfide, Na2S (aq), that is needed to precipitate all the mercury ions in 75.5 mL of 0.100 mol/L, Hg(NO3)2 (aq).

Expert's answer

There is the next equation for this process:

Hg(NO_{3})_{2} + Na_{2}S --> HgS + 2Na(NO_{3})_{2}

For answering this question you need to find amount of Hg(NO_{3})_{2 }in the solution:

0.100 mol per 1 L

x mol per 0.0755 L

x = 0.000755 mol

0.000755 mol of Hg(NO_{3})_{2} can react with 0.000755 mol of Na_{2}S (from equation)

0.000755 0.000755

Hg(NO_{3})_{2} + Na_{2}S --> HgS + 2Na(NO_{3})_{2}

1 1

Now you need to find volume of of 0.0783 mol/L sodium sulfide, that

includes 0.000755 mol of Na_{2}S

0.0783 mol mol per 1 L

0.000755 mol per V

V = 0.009642 L

Hg(NO

For answering this question you need to find amount of Hg(NO

0.100 mol per 1 L

x mol per 0.0755 L

x = 0.000755 mol

0.000755 mol of Hg(NO

0.000755 0.000755

Hg(NO

1 1

Now you need to find volume of of 0.0783 mol/L sodium sulfide, that

includes 0.000755 mol of Na

0.0783 mol mol per 1 L

0.000755 mol per V

V = 0.009642 L

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