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an experiment was performed to investigate the reaction between zn meta…

Question

an experiment was performed to investigate the reaction between zn metal and ni²+ (aq)
at different concentrations. because the ni²+ (aq) ion is green, the extent of the reaction
ni²+ (aq) were prepared by dissolving nicl₂·6h₂o (molar mass 240 g/mol) in water.
the absorbance of each solution was measured; the results are shown both in the table
below and in the following plot of the absorbance data.
what is the expected absorbance of a standard solution made by dissolving 0.0070 mol of
nicl₂·6h₂o in water to make 100. ml

Explanation:

Step1: Calculate the concentration of \(Ni^{2+}\)

The molar mass of \(NiCl_{2}\cdot6H_{2}O\) is \(240\ g/mol\). Given \(n = 0.0070\ mol\) and \(V=100\ mL = 0.1\ L\).
Using the formula \(c=\frac{n}{V}\), we have \(c=\frac{0.0070\ mol}{0.1\ L}=0.070\ M\).

Step2: Determine the relationship between concentration and absorbance

From the data in the table, we can assume a linear relationship (Beer - Lambert law \(A = \epsilon lc\), where \(\epsilon l\) is a constant).
Take two points from the table, say \((c_1 = 0.020\ M,A_1 = 0.12)\) and \((c_2 = 0.040\ M,A_2 = 0.25)\). The slope \(k=\frac{A_2 - A_1}{c_2 - c_1}=\frac{0.25 - 0.12}{0.040 - 0.020}=\frac{0.13}{0.020}=6.5\).

Step3: Calculate the absorbance for \(c = 0.070\ M\)

Using \(A=kc\), substitute \(c = 0.070\ M\) and \(k = 6.5\).
\(A=6.5\times0.070 = 0.455\approx0.45\)

Answer:

D. \(0.45\)