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).
Rather than finding the chromate concentration first and scaling it afterward, it helps to build the chromium-based calibration directly by adjusting the slope of the line up front.
The molecular weight of chromate is $M(CrO_4^{2-}) = 52 + 4(16) = 116$ g/mol, and chromium makes up a fraction $\frac{52}{116} = 0.44828$ of that mass.
Since the original calibration line is $y = 0.8003x + 0.0055$ with $x$ in mg/L of $CrO_4^{2-}$, we can express $x$ in terms of the equivalent Cr concentration $x_{Cr} = 0.44828x$, so $x = \dfrac{x_{Cr}}{0.44828}$. Substituting into the calibration equation: $$0.35 = 0.8003\left(\frac{x_{Cr}}{0.44828}\right) + 0.0055$$
The result agrees with the two-step method, since both simply rescale the same linear relationship in a different order.
\[\boxed{C_{Cr} \approx 0.19\ mg/L}\]
