For preparation of 1000 mL of 0.1 N potassium dichromate solution (0.1 N K\(_2\)Cr\(_2\)O\(_7\), atomic mass of K = 39, Cr = 52, O = 16), the amount of analytical grade potassium dichromate required is:
Show Hint
Remember the standard chemistry values for Walkley-Black organic carbon estimation:
- Molecular weight of \(\text{K}_2\text{Cr}_2\text{O}_7\) = 29
- Equivalent weight in acid = 40.
- Weight for 1 N solution = 40 g/L.
- Weight for 0.1 N solution = 9 g/L.
Step 1: Understanding the Concept:
To prepare a normal solution (\(N\)), we must determine the equivalent weight of the solute.
The equivalent weight depends on the chemical reaction and the number of electrons transferred per molecule during a redox reaction. Key Formula or Approach:
The mass of solute (\(W\)) required to prepare a solution of a given normality is:
\[ W = \text{Normality (N)} \times \text{Equivalent Weight (Eq. Wt.)} \times \text{Volume (L)} \]
\[ \text{Equivalent Weight} = \frac{\text{Molecular Weight}}{\text{Valency factor (n-factor)}} \]
Step 2: Detailed Explanation:
Let's perform the calculations step-by-step:
Calculate the molecular weight of potassium dichromate (\(\text{K}_2\text{Cr}_2\text{O}_7\)):
\[ \text{M. Wt.} = (2 \times 39) + (2 \times 52) + (7 \times 16) \]
\[ \text{M. Wt.} = 78 + 104 + 112 = 294 \text{ g/mol} \]
Determine the equivalent weight of \(\text{K}_2\text{Cr}_2\text{O}_7\):
In an acidic medium, the dichromate ion (\(\text{Cr}_2\text{O}_7^{2-}\)) is reduced to chromium ions (\(\text{Cr}^{3+}\)):
\[ \text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 6e^- \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O} \]
Since each molecule accepts 6 electrons, the valency factor (\(n\)-factor) is 6:
\[ \text{Equivalent Weight} = \frac{294}{6} = 40 \text{ g} \]
Calculate the weight required for 1000 mL (1 L) of 0.1 N solution:
\[ W = 0.1 \text{ N} \times 40 \text{ g} \times 1 \text{ L} = 9 \text{ g} \]
Step 3: Final Answer:
The amount of analytical grade potassium dichromate required is 9 grams.
