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Spectrophotometer was used to measure the_gbsorbance of this transition metal complex at various concentrations. This data was plotted with Absorbance (A) on the Y-...

Question

Spectrophotometer was used to measure the_gbsorbance of this transition metal complex at various concentrations. This data was plotted with Absorbance (A) on the Y-axis and concentration (c) in mol dm On the x-axis: Thc resulting graph was lincar and gave the following equation: y = 243x The solution follows Beer-Lambert's Law where A = ec/ and the path length of the cuvette used is 1 5cm_ Use the information provided to answer the following questions. A sample of the transition metal compl

spectrophotometer was used to measure the_gbsorbance of this transition metal complex at various concentrations. This data was plotted with Absorbance (A) on the Y-axis and concentration (c) in mol dm On the x-axis: Thc resulting graph was lincar and gave the following equation: y = 243x The solution follows Beer-Lambert's Law where A = ec/ and the path length of the cuvette used is 1 5cm_ Use the information provided to answer the following questions. A sample of the transition metal complex had an absorbance of 0.489_ What is the concentration? AnsWer mol di The transition metal complex is formed according to the following equation ligand X(ligand) If the equilibrium constant for this transition mctal complex is 495 and solution had the following cquilibrium concentrations. [X ]=4x 10 [ligand] = 6x 10 Then what is the equilibrium concentration of [X(ligand) ]? Answer mol dm



Answers

In case study $3,$ we showed how colorimetry can be used to follow the oxidation of $\left[\mathrm{C}_{2} \mathrm{O}_{4}\right]^{2-} \mathrm{by}$ $\left[\mathrm{MnO}_{4}\right]^{-}$ ions in acidic aqueous solution. The data shown in Figure 11.8 refer to an experiment in which $2.0 \mathrm{cm}^{3}$ of aqueous $\mathrm{KMnO}_{4}\left(4.0 \times 10^{-4} \mathrm{moldm}^{-3}\right)$ are mixed with $1.0 \mathrm{cm}^{3}$ aqueous sulfuric acid solution that contains $\left[\mathrm{C}_{2} \mathrm{O}_{4}\right]^{2-}$ ions $\left(x \mathrm{moldm}^{-3}\right)$ in a cuvette of path length $1.0 \mathrm{cm}$. (a) Rationalize the shape of the curve shown in Figure $11.8 .$ (b) The initial absorbance (time $=0 \mathrm{s}$ ) is 0.65 Determine the molar extinction coefficient of $\mathrm{KMnO}_{4} \cdot(\mathrm{c})$ If the final absorbance is 0.14, calculate the concentration of $\left[\mathrm{MnO}_{4}\right]^{-}$ ions remaining in solution at the end of the reaction. (d) How many moles of $\left[\mathrm{MnO}_{4}\right]^{-}$ are used during the reaction? How many moles of $\left[\mathrm{C}_{2} \mathrm{O}_{4}\right]^{2-}$ were present initially (see equation 11.6 )? Hence determine the concentration of $\left[\mathbf{C}_{2} \mathbf{O}_{4}\right]^{2-}$ present in the original $1.0 \mathrm{cm}^{3}$ sample of oxalate ions.

Solution for the above problem here. Observance here. Maybe not. Observance by capital verses. We have tracer concentration bus to but through origin and point observance is equal to yeah, 0.9. Concentration is equal to three Mike 0 g Polly to and here c is equal. Do. Okay, okay. Care is equal to see. Divided by a is equal to three. We have concentration. Zero point nineties Observance is equal to We have to believe of gay as a three point 33 No, for a circumstance, aeration is equal to three three point 33 observance and this is a collaboration. The observance is equal to 0.18 Concentration is equal to 3.33 Multiply 0.1. It is equal to 0.6 microgram. Well, later it's amount off. Dye injected. Is it well too? 0.6 kill big soon into one liter. Divided by dozen. Cubic. Soon in two five. My program divided by later into 10 rest of the power tree microgram divided by when my program is equal to three. My program as you total amount Blood is volume in later and here concentration is equal to 0.6 micro ground poorly too, is equal. Do three microgram divided by the volume in later and here, William in Later as you could do quiet nature.

So trans mittens and absorbent are inversely related, so we can calculate the concentrations of cobalt by taking their corresponding Mahler extinction coefficient values and their wavelengths. So we have C is equal to a over e l so cold, but with a wavelength of 510 under meters, we have concentration is equal to 9.446 Divide that by 36400 multiplied by 9.1 That's equal to 1.22 times 10 to the minus. Five more Podesto meats are cute. Next we have 656 nanometers is equal to not point 446 Do I do that by 1 to 40 times 1.0? That's equal to 3.596 times 10 to the minus four most protesters meters. Next we have a nickel to look at. So we've got nickel with a wavelength of 5 10 nanometers. So we have concentration is equal to 9.3 to 6. Invited by 45 to 0, multiplied by 1.0 the length 4 to 5.905 times 10 to the minus five malls decimates a cubed. Next we have neck all the way, but 656 nanometers concentration is able to not 0.3 to 6 by that by 17500 multiplied by 1.0. That gives us a body of 1.862 times 10 to the minus five malls Podesta Meter Cube.

We're given the equation defined absorb INTs and we're told that absorb INTs equals epsilon times L A. Times C epsilon is the molar extinction coefficient. L is the path length in centimeters and C is concentration and polarity, and these units are very important. So the first thing that we need to do is convert our units of milligrams times milliliters in verse into molar ity. So we know we'll start with our first ratio. One milligram for everyone. The leader. You know what everyone? Graham. There are 1000 milligrams and everyone leader. There's 1000 to middle leaders and everyone mall there 70,800 grants. But when you multiply this out and do our dimensional analysis, we'll find that our concentration is 5.6179 times 10 to the negative fifth Moeller. And now we have all the information to plug into our observance equation. Get that are absorbent sequel to 15,000 timeto. One times are concentration and we'll get that are absorb INTs is 0.843 were also given another equation to find the absorb INTs with respect to light intensity. And this equation tells us that absorb INTs is equal to log base. 10. I not divided by I so we can rearrange this equation a little bit for us to use it. We'll get 10 raised to a is equal toe. I not divided by I. Then if we rearranges again, we'll get I divided by I not is equal to 1/10 raise to the A and now we have the information to find the percentage of light we'll get that I divided by I not is equal to one divided by 10 raised to our absorb INTs. We just found 100.843 and we'll get that our decimal this 0.1435 That means that the percent of incident light is 14.35%.

Going on with electric chemistry here where essentially just looking at the movements of electrons throughout our system. First thing we're doing is calculating in East Cell Valley, where we have the style is e cathode reduction. Subtract e honored oxidation s. I'll do not 0.446 bolts. Subtract 9.242 volts. Eso is equal to not 0.204 vaults into the relationship between some potential and the free energy change. Delta G equals negative on F E South, where n is two moles. F is 96385 Sl is not 963850.204 so we can solve a Delta G by plugging in. All our numbers were dealt cheese negative 3.93 times 10 to the four jewels. Following this, we can take a look at the general formula announced equation that it's Esau particular to me. No, the track no point No. 591 divided by and multiplied by log Cute. Thank you is C R for two minus divided by 80 plus We have our products over our reactant. So we have moles That is too sweet. Plug in our values. So we have now have SL was not 0.242 because you calculated Dino struck No point no. 591 find by two of log cute So we're giving concentrations to plug in for C R 04 to minus and 80 plus Then we can solve for ESA that is not 0.352 volts Now, to find the concentration we can use announced equation where we have a lot of known information here we're solving for the c r. 04 to minors. Concentration that is are unknown. And so what we have is not 25 several four volts equal to 9.242 volts. Subtract no point no 591 bye bye to log C R 04 to minus divided by 1.0 small pelita. We then solve this concentration that's missing that is equal to 1.36 times 10 to the minus nine miles Paletta. Lastly, we're calculating the solidarity product of 80 cr four. So given the announced equation again, we have the equilibrium easily call sero. So what we have is no point, no no faults. It's equal to not 0.0.242 vaults. Subtract no point No. 592 volts, divided by two Log A G plus squad CR four to minus divided by a G To see our of four, where we saw for a G Plus squad CR four to minus start is equal to 10 to 8.1895 That is equal to 1.55 times 10 to the eighth on 1.55 times 10 to the eight is Thesis on Ability product for a G two c r 04


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