Calculate the concentration in mol per litre of ammonium ions in a solution solution prepared by mixing 360.0mL of 0.250 mol per litre ammonium sulphate solution with 675.0mL of 1.20mol per litre ammonium nitrate solution. Assume solution volumes are additive.
I got 0.870 mol per Litre...
I don't think it's correct....
mols (NH4)2SO4 = LxM = 0.360L x 0.250M =??
mols NH4NO3 = LxM = 0.675 L x 1.20 M = ??
M NH4^+ of final solution = mol/L
total mols NH4^+ = 2 x mols of (NH4)2SO4 + mols NH4NO3 = ??
total volume = 0.350L + 0.675 L = ??
To calculate the concentration of ammonium ions in the final solution, you need to perform a series of calculations.
First, calculate the number of moles of ammonium sulfate ((NH4)2SO4) in the 360.0 mL solution.
mols (NH4)2SO4 = L x M = 0.360 L x 0.250 M = 0.09 moles
Next, calculate the number of moles of ammonium nitrate (NH4NO3) in the 675.0 mL solution.
mols NH4NO3 = L x M = 0.675 L x 1.20 M = 0.81 moles
Now, calculate the total moles of ammonium ions in the final solution.
total mols NH4+ = 2 x mols of (NH4)2SO4 + mols NH4NO3
total mols NH4+ = 2 x 0.09 moles + 0.81 moles = 0.99 moles
Finally, calculate the concentration of ammonium ions in the final solution.
total volume = 0.360 L + 0.675 L = 1.035 L
concentration of NH4+ = total mols NH4+ / total volume
concentration of NH4+ = 0.99 moles / 1.035 L = 0.955 moles per litre
Therefore, the concentration of ammonium ions in the final solution is 0.955 mol/L, which is different from your initial calculation of 0.870 mol/L.