To calculate the equilibrium constant \( K \) for the hydrolysis of antimony trichloride (SbCl3), we first consider the hydrolysis reaction:
\[ \text{SbCl}_3 + 3 \text{H}_2\text{O} \rightleftharpoons \text{Sb(OH)}_3 + 3 \text{HCl} \]
In this reaction, one mole of SbCl3 reacts with three moles of water to produce one mole of Sb(OH)3 and three moles of HCl.
The equilibrium expression for this reaction can be written as:
\[ K = \frac{[\text{Sb(OH)}_3][\text{HCl}]^3}{[\text{SbCl}_3]} \]
From the data given:
- Concentration of SbCl3 = 0.046 M
- Concentration of HCl = 1.7 M
To find the concentration of Sb(OH)3 at equilibrium, we note that when one mole of SbCl3 reacts, it produces one mole of Sb(OH)3. Therefore, the concentration of Sb(OH)3 is equal to the initial concentration of SbCl3 minus the concentration that remains at equilibrium (which is the difference). Assuming complete reaction, the concentration of Sb(OH)3 will be equal to the initial concentration of SbCl3 converted to it:
\[ [\text{Sb(OH)}_3] \approx [\text{SbCl}3]{initial} - [\text{SbCl}3]{equilibrium} = 0 - 0.046 = 0.046 \text{ M} ; \text{(theoretically, as most reactants are consumed)} \]
Thus, we need to use the concentration changes based on the stoichiometry of the reaction:
At equilibrium, we can approximate [Sb(OH)3] as being equal to the amount of SbCl3 that has reacted, which corresponds to the maximum formed before reaching the equilibrium concentration.
Now, substituting these concentrations into the equilibrium expression:
\[ K = \frac{(0.046)(1.7)^3}{0.046} \]
Notice that the concentration of Sb(OH)3 cancels out:
\[ K = (1.7)^3 \]
Calculating \( (1.7)^3 \):
\[ (1.7)^3 = 4.913 \]
Rounding to two significant figures:
\[ K \approx 4.9 \]
Thus, the equilibrium constant \( K \) for the hydrolysis of antimony trichloride is:
\[ \boxed{4.9} \]
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