Question

CHEMISTRY HELP NEEDED!! How do I do this problem? can someone show me a step-by-step tutorial? I will vote and answer your response, thanks in advance!

Standard Addition Method Problem Often when doing analytical work, we need to spike our sample because of matrix effects that impact our measurements. This involves making a series of solutions that include increasing amounts of a standard solution of known concentration along with a set amount of the unknown solution, diluting to a mark – i.e. using a volumetric flask - then taking measurements. (Probably absorbance here, we can add thiocyanate to turn the iron red). For instance – we might want to determine the concentration of iron in a water sample (cx). We will use 10.00 ml of our unknown sample (V). We will make solutions with 0.00, 5.00, 10.00, 15.00, and 20.00 ml of standard solution. Our standard solution is 11.1 ppm Fe3* (c). Sconcentration We make dilutions of 50.00 mL (VE). We take measurements and they were: Standard Addition 0.00 5.00 10.00 | 15.00 | 20.00 Signal 0.240 0.437 0.621 0.809 1.009 We want to make a plot of the dependent variable vs. the independent variable. (Which is which?) Plot this using excel and use the functions to give you a slope and an intercept. The slope and intercept have the following information: Slope -- No Intercept-KVAR k is a proportionality constant. You now have two equations and can solve for what you want, which is concentration of our unknown sample.

Answer 1

In this method of standard addition, the dependent variable is absorbance or signal and independent variable is the concentration of standard addition

if Cx is the concentration of unknown,then Cx/(Cx+Cs)=signal x/signl(x+s)

the plot between signal and absorbance ,hence ,is linear and can be extrapolated to get the value of Cx at no std addition

concentration of standard addition=volume of standard addition*Cs/Vt

Cs=11.1ppm=11.1mg/L

concentration of standard addition(mg/L)	signal
(0.00L *11.1mg/L)/0.05L=0	0.240
(0.005L *11.1mg/L)/0.05L=1.11	0.437
(0.0010L *11.1mg/L)/0.05L=0.222	0.624
(0.0015L *11.1mg/L)/0.05L=0.333	0.809
(0.0020L *11.1mg/L)/0.05L=0.444	1.009

slope=1.721L/mg=KCs/Vt

k=1.721*Vt/Cs=(1.721L/mg)*0.05L/11.1mg/L=0.00775

intercept=0.242=kVx*Cx/Vt

Cx=0.242*Vt/kVx*=0.242*0.05L/ (0.00775L^3/mg^2 )*0.01L=156 .129

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