Question

The concentration of copper in a sample of sea water is determined by anodic stripping voltammetry using the method of standard additions. The analysis of a 50.0-mL sample gives a peak current of 0.886 mA. After adding a 5.00-mL spike of 10.0 mg/L Cu2+, the peak current increases to 2.52 mA. Calculate the mg/L copper in the sample of sea water.

Answer 1

To calculate the mg/L copper in the sample of sea water, we can use the method of standard additions. The concentration is found to be 40.0 mg/L.

Step-by-step explanation:

To calculate the mg/L copper in the sample of sea water, we can use the method of standard additions. The peak current increases from 0.886 mA to 2.52 mA after adding a 5.00-mL spike of 10.0 mg/L Cu2+.

The increase in current is directly proportional to the concentration of copper in the sample. Using the formula:

(C2 - C1) = (I2 - I1) * (V2 / V1)

Where C2 is the concentration of copper in the sample, C1 is the concentration of copper in the spike solution, I2 is the peak current after adding the spike, I1 is the peak current before adding the spike, V2 is the volume of the sample after adding the spike, and V1 is the volume of the sample before adding the spike.

Plugging in the values:

(C2 - 10.0) = (2.52 - 0.886) * (50.0 / 5.00)

Simplifying the equation gives C2 = (2.52 - 0.886) * (50.0 / 5.00) + 10.0 = 40.0 mg/L.