Question

A detergent containing an unknown concentration of phosphate is analyzed using the standard addition method. Initially, this solution emitted an instrumental signal of 4.78 μA. An aliquot of 10.0 ml of 3.12 M phosphate solution was added to the 80.0 ml sample. After this addition, a signal of 9.77 μA was measured. Calculate the initial concentration of phosphate in the original sample of detergent.

Answer 1

Let unknown concentration of phosphate = C

For the addition,

volume of sample = 80.0 mL

moles of phosphate already present = (unknown concentration of phosphate) * (volume of sample)

moles of phosphate already present = (C) * (80.0 mL)

moles of phosphate already present = 80C mmol

concentration of added solution = 3.12 M

volume of added solution = 10.0 mL

moles of phosphate added = (concentration of added solution) * (volume of added solution)

moles of phosphate added = (3.12 M) * (10.0 mL)

moles of phosphate added = 31.2 mmol

Total moles of phosphate = (moles of phosphate already present) + (moles of phosphate added)

Total moles of phosphate = (80C mmol) + (31.2 mmol)

Total moles of phosphate = (80C + 31.2) mmol

Total volume = (volume of sample) + (volume of added solution)

Total volume = (80.0 mL) + (10.0 mL)

Total volume = 90.0 mL

Final phosphate concentration = (Total moles of phosphate) / (Total volume)

Final phosphate concentration = [(80C + 31.2) mmol] / (90.0 mL)

Final phosphate concentration = (0.89C + 0.347) M

The signal produced is proportional to the phosphate concentration.

Mathematically, A1 / C1 = A2 / C2

where A1 = initial signal = 4.78 uA

C1 = unknown concentration of phosphate = C

A2 = final signal = 9.77 uA

C2 = final phosphate concentration = (0.89C + 0.347) M

Substituting the values,

(4.78 uA) / (C) = (9.77 uA) / (0.89C + 0.347)

4.78 * (0.89C + 0.347) = 9.77 C

Solving for C, C = 0.300 M

initial concentration of phosphate in the original sample = 0.300 M