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A student added 50.0 ml of an naoh solution to 100.0 ml of 0.400 m hcl. The solution was then treated with an excess of aqueous chromium(iii) nitrate, resulting in formation of 2.06 g of precipitate. Determine the concentration of the naoh solution.

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Answer is: concentration of the sodium hydroxide solution is 2M.

Balanced chemical reaction:

3NaOH(aq) + Cr(NO₃)₃(aq) → Cr(OH)₃(s) + 3NaNO₃(aq).

m(Cr(OH)₃) = 2.06 g.

n(Cr(OH)₃) =2.06 g ÷ 103 g/mol.

n(Cr(OH)₃) = 0.02 mol.

From reaction: n(Cr(OH)₃) : n(NaOH) = 1 : 3.

n(NaOH) = 0.06 mol.

From chemical reaction: NaOH(aq) + HCl(aq) → NaCl(aq) + H₂O(l):

n(HCl) = n(NaOH) = 0.1 L · 0.4 M = 0.04 mol.

n(NaOH) = 0.06 mol + 0.04 mol = 0.1 mol.

c(NaOH) = 0.1 mol ÷ 0.05 L = 2 mol/L.


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