The given question is incomplete, the complete question is:
Suppose 1.27 g of potassium iodide is dissolved in 100. mL of a 44.0 m M aqueous solution of silver nitrate. Calculate the final molarity of iodide anion in the solution. You can assume the volume of the solution doesn't change when the potassium iodide is dissolved in it. Round your answer to 3 significant digits.
Answer:
The correct answer is 0.0325 M.
Explanation:
The mass of potassium iodide or KI mentioned in the question is 1.27 grams, the molar mass of KI is 166 g/mol. The formula for determining the no of moles of the substance is mass/molar mass. Thus, the moles of KI in 1.27 grams will be,
= 1.27g / 166 g/mol = 0.00765 moles.
KI = K⁺ + I⁻
Therefore, the moles of KI will be equivalent to moles of iodide anion, that is, 0.00765 moles.
The moles of silver nitrate or AgNo3 in the solution can be determined by using the formula, molarity (M) * volume in liters. The molarity of silver nitrate given in the question is 44 mM and the volume used is 100 ml or 100/1000 L. Now putting the values we get,
= (44 M/1000) * (100 L/1000) = 0.0044 moles
The moles of silver nitrate is equivalent to moles of silver ion, which is further equivalent to the moles of iodide ion that has taken part in precipitation = 0.0044 moles.
The moles left of I⁻ in the solution will be,
0.00765 - 0.0044 = 0.00325
Now, the final molarity of iodide ion in the solution will be,
= moles/volume in liters
= 0.00325 moles / 0.100 L = 0.0325 M
