Answer:
The molarity of the nitrate ion = 0.25 M
Explanation:
Step 1: Data given
Volume of 0.25 M K2Cr2O7 = 100 mL = 0.1 L
Volume of 0.25 M Pb(NO3)2 =100 mL = 0.1 L
Step 2: The balanced equation
Pb(NO3)2(aq) + K2Cr2O7(aq) → PbCr2O7(s) + 2KNO3(aq)
PbCr2O7(s) + 2KNO3(aq) → PbCr2O7(s) + 2K+ + 2NO3-
Step 3: Calculate moles of K2Cr2O7
Moles K2Cr2O7 = Molarity K2Cr2O7 * volume
Moles K2Cr2O7 = 0.25M * 0.1 L
Moles K2Cr2O7 = 0.025 moles
Step 4: Calculate moles of Pb(NO3)2
Moles Pb(NO3)2 = 0.25 M * 0.1 L
Moles Pb(NO3)2 = 0.025 moles
Step 5: Calculate limiting reactant
For 1 mol Pb(NO3)2 consumed, we need 1 mol of K2Cr2O7, to produce 1 mol of PbCr2O7 and 2 moles of KNO3
Both reactant will be completely consumed. since there will react 0.025 moles of both.
Step 6: Calculate moles of KNO3
For 1 mol Pb(NO3)2 consumed, we need 1 mol of K2Cr2O7, to produce 1 mol of PbCr2O7 and 2 moles of KNO3
For 0.025 mol Pb(NO3)2, we have 2*0.025 = 0.050 moles of KNO3
Step 7: Calculate moles of nitrate ion
2 moles KNO3 will dissociate in 2 moles of K+ and 2 moles of NO3-
For 0.050 moles KNO3, we'll have 0.050 moles of NO3-
Step 8: Calculate molarity of NO3- ion
Molarity NO3- = moles NO3- / volume
Molarity NO3- = 0.050 moles/ 0.2L
Molarity NO3- = 0.25 M
The molarity of the nitrate ion = 0.25 M