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------------------------------------------------ A 25.0 mL sample containing Cu²⁺ gave an instrument signal of 23.6 units (corrected for a blank). When exactly 0.500 mL of 0.0287 M Cu(NO₃)₂ was added to the solution, the signal increased to 37.9 units. Calculate the molar concentration of Cu²⁺ assuming that the signal was directly proportional to the analyte concentration.

To find the molar concentration of Cu²+ in the original solution, we calculate the moles added from the standard, find the proportionality constant (k), and use it to determine the moles of Cu²+ present in the 25.0 mL sample. Then, we divide the moles of Cu²+ by the sample volume in liters to find the concentration.

Explanation:

To calculate the molar concentration of Cu²+ in the original 25.0 mL sample, we use the direct proportionality of the signal to the concentration of the analyte.

Original signal for Cu²+ in the 25.0 mL sample: 23.6 units
Signal after adding 0.500 mL of 0.0287 M Cu(NO₃)2: 37.9 units

The increase in signal corresponds to the amount of Cu²+ added from the 0.500 mL of 0.0287 M Cu(NO₃)2 solution. First, calculate the moles of Cu²+ added:

Moles of Cu²+ added = Volume * Concentration = 0.500 mL * 0.0287 M = 1.435e-5 mol

Convert the volume from mL to L by dividing by 1000 (0.500 mL = 0.0005 L).

The proportionality factor (k) can be determined using the increase in signal (37.9 - 23.6 = 14.3 units) and the moles of Cu²+ added.

k = ΔSignal / Moles added = 14.3 units / 1.435e-5 mol

Using the proportionality factor, calculate the moles of Cu²+ in the original 25.0 mL sample:

Moles of Cu²+ in original sample = Original signal / k

Finally, obtain the molar concentration of Cu²+ by dividing the moles of Cu²+ by the original volume (in liters).

Concentration of Cu²+ = Moles of Cu²+ / 0.025 L