To solve this problem, we can use the principles of stoichiometry and the concept of molarity to find the number of milliliters (mL) of potassium iodide (KI) solution needed.
Step 1: Write the balanced chemical equation for the reaction between sodium dichromate (Na2Cr2O7) and potassium iodide (KI):
Na2Cr2O7 + 3KI -> 2NaI + Cr2O3 + KIO3
According to the balanced equation, the molar ratio between Na2Cr2O7 and KI is 1:3. This means that for every 1 mole of Na2Cr2O7, we need 3 moles of KI.
Step 2: Calculate the number of moles of Na2Cr2O7:
Given mass of Na2Cr2O7 = 0.125 g
Molar mass of Na2Cr2O7 = 2(23) + 2(52) + 7(16) = 294 g/mol
Number of moles of Na2Cr2O7 = Given mass / Molar mass
= 0.125 g / 294 g/mol
≈ 0.000425 moles
Step 3: Determine the number of moles of KI required:
From the balanced equation, we know that the molar ratio between Na2Cr2O7 and KI is 1:3. Therefore, the number of moles of KI required will be three times the moles of Na2Cr2O7.
Number of moles of KI = 3 * Number of moles of Na2Cr2O7
= 3 * 0.000425 moles
≈ 0.001275 moles
Step 4: Calculate the volume of the KI solution:
Given concentration of KI solution = 0.025 M (Molarity is measured in moles per liter, or mol/L)
Molarity (M) = Moles of solute / Volume of solution (in liters)
Rearranging the equation, Volume of solution (in liters) = Moles of solute / Molarity
Volume of KI solution (in liters) = 0.001275 moles / 0.025 M
= 0.051 L (since 1 L = 1000 mL)
Step 5: Convert the volume from liters to milliliters:
Volume of KI solution (in mL) = 0.051 L * 1000 mL/L
= 51 mL
Therefore, approximately 51 mL of the potassium iodide solution will be needed to react with 0.125 g of sodium dichromate.