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Pure solid NaH2PO4 is dissolved in distilled water, making 100.0ml of solution.

ID: 782341 • Letter: P

Question

Pure solid NaH2PO4 is dissolved in distilled water, making 100.0ml of solution. 10 ml of this solution is diluted to 100.00 ml to prepare the orignal phosphare standard solution. Three working standard solution are made from this by pipetting into 0.8 ml, 1.5 ml and 3.0 ml of the orignal standard solution into 100.0 ml volumetric flasks. Acid and molybdate reagent are added and the solution are diluted to 100.0 ml. phosphate in the soft drink. 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. A 10.0 ml volume of soft drink is treated with acid and reagent and also diluted to 100.00 ml. Its absorbance is also measured. From the following data, plot a calibration line (absorbance vs. concentration), determine its slope and calculate the concentration of phospate in the original soft drink(mM).

Data given:

Mass of NaH2PO4 (mg) -- 536.4

Absorbance standard 1 (0.8 ml) -- 0.1973

Standard 2 (1.5 ml) --- 0.3701

Standard 3 (3.0 ml) --- 0.7402

sample (10.0 ml) -- 0.2230

Please help me with this calculation:

Concentration of orignal phosphate standard (mM)

concentration of standard 1(mM)

concentration of standard 2(mM)

concentration of standard 3(mM)

Slope of calibration line

calculate the concentration of phospate in the original soft drink(mM).

Explanation / Answer

number of milli moles=536.4/120

=4.47 milli moles

concentration of the original solution=4.47*(1000/100)*(10/100)

=4.47 mM

concentration of standard>

=0.03576 mM

concentration of standard two=4.47*1.5/100

=0.06705 mM

concentration of the third standard=4.47*3/100

=0.1341 mM


let the line be y=mx+c

so,

0.1973=0.03576*m+c

and 0.3701= 0.06705m+c

solving , we get the slope m to be 5.522

so slope=5.522

let the concentration be x.so,

0.2230=5.22*x-1.857*10^-4

or x=0.04276 mM

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