Question

A Phosphate in urine can be determined by spectrophotometry. After removing protein from the sample, it is treated with a molybdenum compound to give, ultimately, a deep blue polymolybdate. The absorbance of the blue polymolybdate can be measured at \(650 \mathrm{nm}\) and is directly related to the urine phosphate concentration. A 24 -hour urine sample was collected from a patient; the volume of urine was 1122 mL. The phosphate in a \(1.00 \mathrm{mL}\), portion of the urine sample was converted to the blue polymolybdate and diluted to \(50.00 \mathrm{mL} .\) A calibration curve was prepared using phosphate-containing solutions. (Concentrations are reported in grams of phosphorus (P) per liter of solution.) $$\begin{array}{lc} \text { Solution (mass P/L) } & \begin{array}{c} \text { Absorbance at } 650 \mathrm{nm} \\ \text { in a } 1.0-\mathrm{cm} \text { cell } \end{array} \\ \hline 1.00 \times 10^{-6} \mathrm{g} & 0.230 \\ 2.00 \times 10^{-6} \mathrm{g} & 0.436 \\ 3.00 \times 10^{-6} \mathrm{g} & 0.638 \\ 4.00 \times 10^{-6} \mathrm{g} & 0.848 \\ \text { Urine sample } & 0.518 \\ \hline \end{array}$$ (a) What are the slope and intercept of the calibration curve? (b) What is the mass of phosphorus per liter of urine? (c) What mass of phosphate did the patient excrete in the one-day period?

Short answer

(a) Slope: 0.206, Intercept: 0.024; (b) 1.2 x 10^-4 g/L; (c) 1.3464 x 10^-4 g

Step-by-step solution

Plot Calibration Curve and Calculate Slope and Intercept

Using the given data points for solutions (Concentration vs Absorbance), apply linear regression to find the best fit line. This is done by plotting concentration on the x-axis and absorbance on the y-axis. Using statistical tools or calculator, the linear equation is of the form \( y = mx + b \), where \( m \) is the slope and \( b \) is the intercept.

Determine the Slope and Intercept

Calculate the slope \( m \) as the change in absorbance over the change in concentration and the intercept \( b \). The slope \( m \) is determined as 0.206 and the intercept \( b \) is approximately 0.024.

Calculate the Concentration of Urine Sample

With the absorbance of the urine sample, which is 0.518, use the equation of the line from the calibration curve: \( y = 0.206x + 0.024 \). Rearrange it to solve for \( x \), the concentration of phosphorus in grams per liter. Substitute \( y = 0.518 \) to get \( x = \frac{0.518 - 0.024}{0.206} \approx 2.4 \times 10^{-6} \) g/L.

Adjust for Dilution

Consider that the urine sample was diluted from \(1.00\) mL to \(50.00\) mL, which means the concentration calculated above is for the diluted solution. Multiply the calculated concentration by the dilution factor, which is \(50\) (since \(50.00\) mL was measured from the \(1.00\) mL initial volume). Thus, the concentration of phosphorus in the original urine sample is \(2.4 \times 10^{-6} \times 50 = 1.2 \times 10^{-4} \) g/L.

Calculate Total Mass of Phosphate Excreted

The total volume of urine collected for 24 hours is 1122 mL or 1.122 L. Multiply the concentration of phosphorus in the urine (\(1.2 \times 10^{-4} \) g/L) by the total volume of urine (1.122 L) to find the mass of phosphorus. The calculation yields \(1.3464 \times 10^{-4} \) grams of phosphorus in total excretion for the one-day period.

Key concepts

Phosphate determination

In the realm of analytical chemistry, phosphate determination in a sample often relies on spectrophotometry. This technique involves using the reaction of phosphate with specific reagents to yield a compound that absorbs light at a particular wavelength. For the exercise at hand, phosphate in urine reacts with a molybdenum compound, resulting in a blue polymolybdate complex. This complex is advantageous because it has a specific absorbance at 650 nm, making it easily measurable.
Once the reaction is complete, the absorbance of the sample can be measured. This absorbance relates directly to the concentration of phosphate present, allowing chemists to quantify the amount of phosphate in the urine sample. It's important to ensure that any interfering substances, like proteins, are removed beforehand to maintain accuracy.

By understanding this process, chemists can effectively gauge the phosphate levels in various samples, facilitating diverse applications such as medical diagnostics, environmental monitoring, and research.

Calibration curve

The calibration curve is a fundamental tool in spectrophotometry and is critical for determining unknown concentrations in samples. It is constructed by measuring the absorbance of a set of standard solutions with known concentrations. The absorbance values are plotted on the y-axis against the concentration on the x-axis. These data points should form a straight line if the relationship is linear, which is typical for low concentrations.
In this exercise, a series of phosphate standards were made, and their absorbance was measured at 650 nm. These measurements are then used to plot the calibration curve. The line of best fit is typically derived using linear regression which calculates the slope (m) and the y-intercept (b).

• The slope indicates how much the absorbance changes with concentration.
• The intercept ideally should be close to zero if no other absorptive factors are present.

Understanding how to create and utilize a calibration curve is crucial as it allows for the determination of an unknown sample's concentration by simply measuring its absorbance and referring back to the established line.

Absorbance measurement

Absorbance measurement is at the heart of determining concentrations in spectrophotometric analysis. It is based on the Beer-Lambert Law, which states that absorbance ( A ) is directly proportional to the concentration (c) of the absorbing species in the path length (b) of the light, expressed as A = εcb. In our exercise, measuring the absorbance of the urine sample at 650 nm allows us to trace back to the concentration of phosphate within it. The basics of spectrophotometry include preparing the sample appropriately, ensuring all reactants are mixed thoroughly, and using a spectrophotometer to pass a beam of light through the sample.
Important considerations include the correct calibration of the spectrophotometer and the selection of the appropriate cuvette path length, which is constant in this setup, typically 1 cm.

This method provides a high degree of accuracy and is integral to the determination of concentrations in various fields, from clinical analysis to pharmaceuticals.

Linear regression

Linear regression serves as a statistical approach to find the line of best fit through a series of data points. In spectrophotometry, this technique is essential for constructing calibration curves, allowing us to interpret the relationship between absorbance and concentration.
When constructing this line, the mathematical form used is y = mx + b, where y is the absorbance, x is the concentration, m is the slope, and b is the intercept.

• The slope (m) represents the change in absorbance per unit change in concentration.
• The intercept (b) ideally touches close to the origin in cases where the solution has no absorbance at zero concentration.

In the exercise provided, linear regression was applied to the absorbance values obtained from phosphate standards to derive m = 0.206 and b = 0.024. With these parameters, the absorbance measured from unknown samples can be used to back-calculate their concentrations. Understanding linear regression is vital as it validates the relationship between variables, ensuring reliable concentration analyses across scientific applications.