Quantitative question... please guide me on how to solve this problem.?
A solution is found to contain 10 ppm ammonium ion. How much (in grams) ammounium phosphate (95% pure) is needed to prepare 200ml of the solution. Calculate the Normality of phosphate anion [PO4^(-3)] and ammonium cation [NH4^(+1)] and, pPO4 and pNH4.
thanks you so much...:)
Answer:
10 ppm = 10 mg /L
mass NH4+ = 18 g/mol
10 mg = 0.001 g
0.001 / 18 = 0.000555 moles NH4+
Ammonium phosphate is (NH4)3PO4
Moles PO43- = 0.000555 / 3 = 0.000185
We have 0.00185 moles (NH4)3PO4
molar mass = 148.97 g/mol
0.00185 mol x 148.97 / mol = 0.275 g
95 : 100 = 0.275 : x
x = 0.289 g in 1 L
in 200 mL we need 0.289 / 5 = 0.0579 g
1 eq =148.97 / 3 = 49.6 g
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thanks you so much...:)
Answer:
10 ppm = 10 mg /L
mass NH4+ = 18 g/mol
10 mg = 0.001 g
0.001 / 18 = 0.000555 moles NH4+
Ammonium phosphate is (NH4)3PO4
Moles PO43- = 0.000555 / 3 = 0.000185
We have 0.00185 moles (NH4)3PO4
molar mass = 148.97 g/mol
0.00185 mol x 148.97 / mol = 0.275 g
95 : 100 = 0.275 : x
x = 0.289 g in 1 L
in 200 mL we need 0.289 / 5 = 0.0579 g
1 eq =148.97 / 3 = 49.6 g
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