Weight of beaker 49.24 g Weight of beaker with NaH_2PO_4 49.77 g absorbance of u
ID: 1054117 • Letter: W
Question
Weight of beaker 49.24 g Weight of beaker with NaH_2PO_4 49.77 g absorbance of unknown .298 .281 average A .2895 volume of a cola used 12 ml A's for cola blank .303 .311 Average A .307 A's for COLA 1.74 1.76 Average A 1.75 Calculated conc. Of standard phosphate solution 4.417 mM Absorbance of measured solution. .2895 Concentration of measured solution (from plot) mM Concentration of PHOSPHATE UNKNOWN mM Absorbance of measured Cola solution .307 Concentration of measured Cola solution (from plot) mM Concentration of phosphate in COLA mM Grams of phosphoric acid per 12 oz can g (12 oz = 355 ml)Explanation / Answer
Provided, the equation of the standard graph, Y = 0.1211 X – 0.068
Let’s have a brief look at the graph and equation.
In the graph, Y-axis indicates absorbance and X-axis depicts concentration. That is, according to the concentration, 1 absorbance unit (1 Y = Y) is equal to 0.1211 units on X-axis (concentration) minus 0.068. So, if we have a known value of Y, we can calculate it counterpart on X-axis. Moreover, phosphorous gives a perfectly linear graph. The intercept of -0.068 (it should have been nearly equal to zero) indicates that the error has occurred somewhere in the experiment. A perfect linear graph has an intercept of zero or nearly zero.
Any, using the equation-
Case 1: Y = 0.2985
Or, 0.2985 = 0.1211 X – 0.068
Or, X = (0.2985 + 0.0685) / 0.1211 = 3.03
Hence, the concentration of the unknown = 3.03 mM (the unit is same as used to plot graph)
Case 2: Y = 0.307
Or, 0.307 = 0.1211 X – 0.068
Or, X = (0.307 + 0.0685) / 0.1211 = 3.100
Hence, the concentration of the unknown = 3.10 mM (the unit is same as used to plot graph)
Note: Actual absorbance of cola = (Abs of cola unknown – abs of cola blank). Abs of cola unknown is above 1.0- it indicates that either some error has occurred or the unknown solution needs to be diluted accordingly to get absorbance below 1.0.
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