The calibration graph between absorbance and the concentration of chromate (\(CrO_4^{2-}\)) in a water sample is shown in the figure.
The calibration line shown in the figure has the equation \(y = 0.8003x + 0.0055\), where \(y\) is the absorbance and \(x\) is the concentration of chromate (\(CrO_4^{2-}\), mg/L).
For an absorbance of 0.35 in a water sample, the estimated Cr(VI) concentration is ______ mg/L (rounded off to two decimal places).
Use the atomic weight (g/mol) of Cr and O as 52 and 16, respectively.
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First back-calculate the chromate (CrO4 2-) concentration from the calibration line, then scale it down by the mass fraction of chromium in CrO4 2- (52 out of 116 g/mol) to get Cr(VI).
Step 1: Use the calibration line to find the chromate concentration. The calibration equation is \(y = 0.8003x + 0.0055\), with \(y\) as absorbance. Substituting \(y = 0.35\): \[0.35 = 0.8003x + 0.0055\] \[0.8003x = 0.3445\] \[x = \frac{0.3445}{0.8003} = 0.43046\ mg/L\] This is the concentration expressed as the chromate ion, \(CrO_4^{2-}\), not as chromium itself.
Step 2: Find the molecular weight of \(CrO_4^{2-}\). \[M(CrO_4^{2-}) = 52 + 4\times16 = 52+64 = 116\ g/mol\]
Step 3: Convert the chromate concentration to Cr(VI) concentration. Only the chromium atom is being reported as Cr(VI), so the mass must be scaled by the fraction of the molecular weight that is chromium: \[C_{Cr} = x \times \frac{52}{116} = 0.43046 \times 0.44828 = 0.19296\ mg/L\]
Step 4: Round the answer. Rounding to two decimal places gives \(C_{Cr} \approx 0.19\ mg/L\), which sits within the expected 0.18-0.21 mg/L range.
