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Harman [31]
3 years ago
10

Let us assume that fe(oh)2(s) is completely insoluble, which signifies that the precipitation reaction with naoh(aq) (presented

in the transition) would go to completion. fe2+(aq)+2naoh(aq) → fe(oh)2(s)+2na+(aq) if you had a 0.500 l solution containing 0.0230 m of fe2+(aq), and you wished to add enough 1.29 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.
Chemistry
1 answer:
Yuki888 [10]3 years ago
3 0

17.8 mL NaOH

<em>Step 1.</em> Write the chemical equation

Fe^(2+) + 2NaOH → Fe(OH)2 + 2Na^(+)

<em>Step 2.</em> Calculate the moles of Fe^(2+)

Moles of Fe^(2+) = 500 mL Fe^(2+) × [0.0230 mmol Fe^(2+)]/[1 mL Fe^(2+)]

= 11.50 mmol Fe^(2+)

<em>Step 3.</em> Calculate the moles of NaOH

Moles of NaOH = 11.50 mmol Fe^(2+) × [2 mmol NaOH]/[1 mmol Fe^(2+)]

= 23.00 mmol NaOH

<em>Step 4.</em> Calculate the volume of NaOH

Volume of NaOH = 23.00 mmol NaOH × (1 mL NaOH/1.29 mmol NaOH)

= 17.8 mL NaOH

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