Calculate the concentration of ammonium ions and sulfate ions in the final solution.?

mol of ammonium sulfate = 19.1g / 132.14g/mol = 0.14 mol

the concentration of ammonium sulphate in the stock = 0.14mol / 0.142L = 1.02 M

the concentration of ammonium sulphate in final solution :

M1 V1 = M2 V2

1.02 x 11.31 = M2 x (11.30 + 52.10)

M2 = 0.18 M

concentration of ammonium ions in 0.18 M of (NH4)2SO4 = 2 x 0.18M = 0.36M

concentration of sulphate ions in 0.18 M of (NH4)2SO4 = 1 x 0.18M = 0.18M

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Step a million: convert grams of (NH4)2 SO4 into moles (hint: use the molecular weight) Step 2: comprehend that when it dissolves, it is going to variety 2 NH4+ ions and basically one SO4 -- ion Step 3: calculate the volum (in LITERs) formed by utilising the answer (hint: 12 + 51; then convert from mL to L) Step 4: Calculate the concentration of SO4 -- by utilising dividing the style of mole by utilising the style of Liters (this might desire to be obvious with the aid of fact the units for concentration are mole / L Step 5: are you able to discover an uncomplicated thank you to make your recommendations up the NH4+ concentration? (hint: 2 NH4+ ions variety for one and all SO4 -- ion)