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.

Question

Chemistry

Quentin Howard

22 November, 02:55

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.

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Peter Russell · Answer 1

17.8 mL NaOH

Step 1. Write the chemical equation

Fe^ (2+) + 2NaOH → Fe (OH) 2 + 2Na^ (+)

Step 2. 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+)

Step 3. Calculate the moles of NaOH

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

= 23.00 mmol NaOH

Step 4. Calculate the volume of NaOH

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

= 17.8 mL NaOH