Calculate final molarity of iodide anion in the solution.?
2017-04-02 3:06 pm
Suppose 35.7g of potassium iodide is dissolved in 350.mL of a 0.50M aqueous solution of silver nitrate. You can assume the volume of the solution doesn't change when the potassium iodide is dissolved in it.
回答 (2)
2017-04-02 6:39 pm
KI(aq) + AgNO₃(aq) → AgI(s) + KNO₃(aq)
In reaction, mole ratio KI : AgNO₃ = 1 : 1
Molar mass of KI = (39.0 + 126.9) g/mol = 165.9 g/mol
Initial no. of moles of KI = (35.7 g) / (165.9 g) = 0.215 mol
Initial no. of moles of AgNO₃ = (0.50 mol/L) × (350/1000 L) = 0.175 mol < 0.215 mol
Hence, KI is in excess, and AgNO₃ completely reacts.
No. of moles of AgNO₃ reacted = 0.175 mol
No. of moles of KI reacted = 0.175 mol
No. of moles of KI left unreacted = (0.215 - 0.175) mol = 0.040 mol
1 mole of KI contains 1 mole of I⁻ anion.
No. of moles of I⁻ left unreacted = 0.040 mol
Volume of the final solution = 350 mL = 0.350 L
Molarity of I⁻ anion left unreacted in the final solution = (0.040 mol) / (350 mL) = 0.114 M
In reaction, mole ratio KI : AgNO₃ = 1 : 1
Molar mass of KI = (39.0 + 126.9) g/mol = 165.9 g/mol
Initial no. of moles of KI = (35.7 g) / (165.9 g) = 0.215 mol
Initial no. of moles of AgNO₃ = (0.50 mol/L) × (350/1000 L) = 0.175 mol < 0.215 mol
Hence, KI is in excess, and AgNO₃ completely reacts.
No. of moles of AgNO₃ reacted = 0.175 mol
No. of moles of KI reacted = 0.175 mol
No. of moles of KI left unreacted = (0.215 - 0.175) mol = 0.040 mol
1 mole of KI contains 1 mole of I⁻ anion.
No. of moles of I⁻ left unreacted = 0.040 mol
Volume of the final solution = 350 mL = 0.350 L
Molarity of I⁻ anion left unreacted in the final solution = (0.040 mol) / (350 mL) = 0.114 M
2017-04-02 4:02 pm
Solutions of sodium carbonate and silver nitrate react to form solid silver carbonate and a solution of sodium nitrate. A solution containing 3.30 g of sodium carbonate is mixed with one containing 3.00 g of silver nitrate. How many grams of each of the following compounds are...
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