The amount of Analytical Reagent grade potassium dichromate required for preparation of 1000 mL of potassium dichromate solution (0.1 N K$_2$Cr$_2$O$_7$). The atomic weight of K = 39, Cr = 52, and O = 16, is:
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In redox reactions under acidic conditions, potassium dichromate has a valency factor of 6.
Its equivalent weight is \(49\text{ g}\) (\(294/6\)).
Thus, \(4.9\text{ g}\) is needed for a \(0.1\text{ N}\) solution, which is the standard primary standard reagent used in Walkley-Black soil organic carbon testing.
Step 1: Understanding the Concept:
Preparing standard normal solutions of chemical reagents requires calculating the equivalent weight of the substance.
The equivalent weight depends on the molecular weight of the compound and the number of electrons exchanged per molecule in the specific redox reaction. Key Formula or Approach:
The molecular weight of potassium dichromate (\(K_2Cr_2O_7\)) is calculated as:
\[ \text{MW} = 2(K) + 2(Cr) + 7(O) \]
In an acidic medium (such as during soil organic carbon estimation by the Walkley-Black method), dichromate acts as a strong oxidizing agent, accepting \(6\) electrons per molecule:
\[ Cr_2O_7^{2-} + 14H^+ + 6e^- \rightarrow 2Cr^{3+} + 7H_2O \]
Therefore, the equivalent weight (EW) is calculated as:
\[ \text{EW} = \frac{\text{Molecular Weight}}{6} \]
The mass required to prepare a solution of normality (\(N\)) in volume (\(V\)) in liters is given by:
\[ \text{Mass} = N \times \text{EW} \times V \] Step 2: Detailed Explanation:
Let us calculate each value step-by-step:
Calculate the molecular weight of \(K_2Cr_2O_7\):
\[ \text{MW} = 2(39) + 2(52) + 7(16) = 78 + 104 + 112 = 294\text{ g/mol} \]
Next, calculate the equivalent weight of \(K_2Cr_2O_7\) in an acidic medium:
\[ \text{EW} = \frac{294}{6} = 49\text{ g} \]
To prepare \(1000\text{ mL}\) (which is \(1\text{ Liter}\)) of a \(0.1\text{ N}\) solution:
\[ \text{Mass} = 0.1 \times 49\text{ g} \times 1\text{ L} = 4.9\text{ g} \]
Therefore, the amount of potassium dichromate required is \(4.9\text{ g}\). Step 2: Final Answer:
The amount of potassium dichromate required is 4.9 g.
