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If you had a 0.550 L solution containing 0.0250 M of Cr3+(aq), and you wished to add enough 1.27 M NaOH(aq) to precipitate all of the metal, what is the minimum amount of the NaOH(aq) solution you would need to add? Assume that the NaOH(aq) solution is the only source of OH−(aq) for the precipitation.

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Answer:

0.032 L of NaOH

Explanation:

The reaction is;

Cr^3+(aq) + 3OH^- (aq) -----> Cr(OH)3(s)

Amount of Cr^3+ = 0.550 × 0.0250 = 0.01375 moles

If 1 mole of Cr^3+ reacts with 3 moles of OH^-

0.01375 moles of Cr^3+ reacts with 0.01375 moles × 3/1

= 0.04125 moles of OH^-

Hence we need 0.04125 moles/1.27 M = 0.032 L of NaOH

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