Pure solid NaH 2 PO 4 is dissolved in distilled water, making 100.00 ml of solut

Question

Pure solid NaH2PO4 is dissolved in distilled water, making 100.00 ml of solution. 10.00ml of this solution is diluted to 100.00 ml to prepare the original phosphate standard solution. Three working standard solutions are made from this by pipetting 0.8ml, 1.5ml and 3.0 ml of the original standard solution into 100.00 ml volumetric flasks. Acid and molybdate reagent are added and the solutions are diluted to 100.00 ml.

You may assume that all these absorbance measurements have already been corrected for any blank absorbance.

The absorbance of each is measured in the spectrophotometer.

Mass of NaH2PO4 (mg) 512.2
Absorbance    Standard 1 (0.8 ml)    0.1922
                          Standard 2 (1.5 ml) 0.3604
                          Standard 3 (3.0 ml) 0.7209

Calculate the following (all one question)
a. Concentration of original phosphate standard (mM)

b. Concentration of standard 1 (mM)

c. Concentration of standard 2 (mM)

d. Concentration of standard 3 (mM)

e. slope of calibration line

Explanation / Answer

(a) Concentration of original phosphate standard (mM) = no. of moles = 512.2mg/ 119.98gmol = 4.27mM

(b) Concentration of standard 1 (mM)

Use M1V1 = M2V2

4.27E-3M x 0.0008mL = M2 x 0.1mL

M2 = 3.42E-5M = 0.0342 mM

(c) Concentration of standard 2 (mM)

Use M1V1 = M2V2

4.27E-3M x 0.0015mL = M2 x 0.1mL

M2 = 6.4E-5M = 0.064 mM

(d) Concentration of standard 3 (mM)

Use M1V1 = M2V2

4.27E-3M x 0.003mL = M2 x 0.1mL

M2 = 3.42E-4M = 0.128 mM

(e) slope of calibration line

We know A = eCL

So, slope = A/C

use any standard. For eg standard 3

Absorbance/Conc = 0.7209/0.128 = 5.63 is the slope.

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