Given the reaction: M(H2O)6^2+ + 2OH- > M(H2O)4(OH)2 + 2H2O for each mL of 0.1 M

Question

Given the reaction: M(H2O)6^2+ + 2OH- > M(H2O)4(OH)2 + 2H2O for each mL of 0.1 M solution of the metal ion, how many mL of 1 M NaOH would be required for complete reaction and what volume constitutes an excess of OH? Given the reaction: M(H2O)6^2+ + 2OH- > M(H2O)4(OH)2 + 2H2O for each mL of 0.1 M solution of the metal ion, how many mL of 1 M NaOH would be required for complete reaction and what volume constitutes an excess of OH? for each mL of 0.1 M solution of the metal ion, how many mL of 1 M NaOH would be required for complete reaction and what volume constitutes an excess of OH?

Explanation / Answer

1 metal ion reacts with 2 OH-

Thus, for each 0.1 M solution of metal ion (consider 1 L solution) we would need = (0.1 M x 1 L)2/1 L = 0.2 L of 1 M NaOH

The volume that constituted excess of OH would be greater then 0.2 L of 1 M NaOH

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