Given a #6.174*g# mass of sodium sulfate that is dissolved in a #1*L# volume, what is the concentration in terms of #"parts per million"# with respect to sodium ion?

Question

Noah Aldrich · Accepted Answer

With respect to sodium ion, we have a #"2000 ppm"# concentration.

#"1 ppm"# specifies #1*mg# of solute per litre of solution...... We has #((6.174*g)/(142.04*g*mol^-1))/(1*L)=0.0435*mol*L^-1# with respect to sodium sulfate.... And thus #2xx0.04346*mol*L^-1=8.693xx10^-2*mol*L^-1#, with respect to sodium ion. Why did I double the concentration of sodium sulfate? And thus, with respect to sodium ion, we have a concentration by mass of ........ #8.693xx10^-2*mol*L^-1xx22.99*g*mol^-1=2.000*g*L^-1#. But #"1 ppm"# #=# #1xx10^-3*g*L^-1#, i.e. #1*mg*L^-1#, and thus we have..... #2.000*g*L^-1xx10^3*mg*g^-1=2000*mg*L^-1-=2000# #"ppm"#.

Kai Bowman · Answer

undefined To find the concentration of sodium ions in parts per million (ppm), you need to first calculate the number of moles of sodium sulfate using its molar mass, and then determine the concentration of sodium ions.

1. Calculate the molar mass of sodium sulfate (Na2SO4):
   Molar mass(Na) = 22.99 g/mol
   Molar mass(S) = 32.07 g/mol
   Molar mass(O) = 16.00 g/mol
   Molar mass(Na2SO4) = 22.99*2 + 32.07 + 16.00*4 = 142.04 g/mol

2. Calculate the number of moles of sodium sulfate:
   moles = mass / molar mass
   moles = 6.174 g / 142.04 g/mol ≈ 0.0435 mol

3. Sodium sulfate dissociates into two sodium ions (Na⁺) for every one molecule of sodium sulfate (Na2SO4).
   Therefore, the number of moles of sodium ions is twice the number of moles of sodium sulfate:
   moles of Na⁺ = 2 * 0.0435 mol = 0.087 mol

4. Calculate the concentration of sodium ions in ppm:
   ppm = (moles of solute / volume of solution in liters) * 10^6
   ppm = (0.087 mol / 1 L) * 10^6 = 87,000 ppm

So, the concentration of sodium ions in terms of parts per million (ppm) is 87,000 ppm.