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A sample of solution has a mixture of 0.01 M Aluminum Chloride and 0.05 M Magnesium Chloride. What is the molarity of Silver that is required to precipitate all the chloride ions present in the solution?

Respuesta :

Answer:

Explanation:

AlCl₃ ⇒AlâșÂł + 3Cl⁻Âč

.01M                 3 x .01 M

So .01 M AlCl₃ will give 3 x .01 M Cl⁻Âč

MgCl₂ ⇒ MgâșÂČ + 2Cl⁻Âč

.05 M                   2 x .05M

.05M MgCl₂ will give  2 x .05 M Cl⁻Âč

Total Cl⁻Âč formed = 3 x .01 + 2 x .05 M

= .03 + .1 M

= .13 M

AgCl ⇒ Agâș + Cl⁻Âč

So AgCl required = .13 M . to precipitate all chloride ions.

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