A solution of 0.200M NaOH (14.0 mL) is mixed with 42.4 mL of 0.170M HNO3. Assuming the final volume is the sum of the initial volumes, what is the molarity of Na+ cation and NO3- anion?
I do not know where to start to solved for the molarity of the ions. Please help !
What's the M of the NaOH after mixing and before reaction?
0.200 M NaOH x (14.0/(14.0+42.4) = approx 0.00025. Since there is 1 Na^+ in 1 molecule of NaOH, 0.00025 is the M of the Na^+.
The nitrate is done the same way. You need not worry about the reaction since Na^+ and NO3^- are spectator ions. By the way that 0.00025 is an estimate; you should recalculate more accurately.
To determine the molarity of Na+ cation and NO3- anion in the solution, we need to calculate the number of moles of NaOH and HNO3 that react and then divide by the final volume of the solution.
Step 1: Calculate the moles of NaOH:
Moles of NaOH = Molarity of NaOH × Volume of NaOH solution
= 0.200 mol/L × 0.0140 L
= 0.00280 moles
Step 2: Calculate the moles of HNO3:
Moles of HNO3 = Molarity of HNO3 × Volume of HNO3 solution
= 0.170 mol/L × 0.0424 L
= 0.00719 moles
Step 3: Write the balanced chemical equation for the reaction between NaOH and HNO3:
NaOH + HNO3 → NaNO3 + H2O
Step 4: Determine the limiting reactant:
The limiting reactant is the one that is completely consumed in the reaction. In this case, NaOH and HNO3 react in a 1:1 molar ratio, so the limiting reactant is HNO3 since there are more moles of HNO3 compared to NaOH.
Step 5: Calculate the moles of NaNO3 formed:
Since NaOH and HNO3 react in a 1:1 molar ratio, the moles of NaNO3 formed will be equal to the moles of HNO3:
Moles of NaNO3 formed = Moles of HNO3
= 0.00719 moles
Step 6: Calculate the final volume of the solution:
Final volume = Volume of NaOH solution + Volume of HNO3 solution
= 0.0140 L + 0.0424 L
= 0.0564 L
Step 7: Calculate the molarity of Na+ cation:
Molarity of Na+ cation = Moles of NaNO3 formed / Final volume of solution
= 0.00719 moles / 0.0564 L
≈ 0.127 mol/L
Step 8: Calculate the molarity of NO3- anion:
Molarity of NO3- anion = Moles of NaNO3 formed / Final volume of solution
= 0.00719 moles / 0.0564 L
≈ 0.127 mol/L
Therefore, the molarity of Na+ cation and NO3- anion in the solution is approximately 0.127 M.