The mobility of a divalent cation in water is 8 × 10-8 m2 V-1 s-1. The effective radius of the ion is (viscosity of water = 1 cP : c = 1.6 × 10-19 C)
212 pm
Effective Radius Calculation from Mobility
The question asks us to find the effective radius of a divalent cation in water, given its mobility, the viscosity of water, and the elementary charge.
We are given the following information:
- Mobility of the divalent cation, \( \mu = 8 \times 10^{-8} \) m\(^2\) V\(^{-1}\) s\(^{-1}\)
- The ion is divalent, meaning its charge magnitude is \( z = 2 \) times the elementary charge.
- Viscosity of water, \( \eta = 1 \) cP. We need to convert this to Pa s (or N s m\(^{-2}\)). \( 1 \) cP \( = 10^{-2} \) P \( = 10^{-2} \) g cm\(^{-1}\) s\(^{-1} \). The conversion factor is \( 1 \) P \( = 0.1 \) Pa s, so \( 1 \) cP \( = 0.1 \times 10^{-2} \) Pa s \( = 10^{-3} \) Pa s.
- Elementary charge, \( q = 1.6 \times 10^{-19} \) C.
Relating Mobility and Radius
The mobility of an ion in a solution is related to its charge, the viscosity of the solvent, and the effective radius of the ion by the Stokes-Einstein relation for ionic mobility. The formula is:
\[ \mu = \frac{zq}{6\pi\eta r} \]where:
- \( \mu \) is the mobility
- \( z \) is the valency (magnitude of the charge number)
- \( q \) is the elementary charge
- \( \eta \) is the viscosity of the solvent
- \( r \) is the effective radius of the ion
Calculating the Effective Radius
We want to find the effective radius \( r \). We can rearrange the formula to solve for \( r \):
\[ r = \frac{zq}{6\pi\eta\mu} \]Now, we can substitute the given values into this equation:
- \( z = 2 \)
- \( q = 1.6 \times 10^{-19} \) C
- \( \eta = 10^{-3} \) Pa s
- \( \mu = 8 \times 10^{-8} \) m\(^2\) V\(^{-1}\) s\(^{-1}\)
Substituting these values, we get:
\[ r = \frac{2 \times 1.6 \times 10^{-19} \text{ C}}{6\pi \times (10^{-3} \text{ Pa s}) \times (8 \times 10^{-8} \text{ m}^2 \text{ V}^{-1} \text{ s}^{-1})} \] \[ r = \frac{3.2 \times 10^{-19} \text{ C}}{48\pi \times 10^{-11} \text{ Pa s m}^2 \text{ V}^{-1} \text{ s}^{-1}} \]Recall that 1 Pa s = 1 N s m\(^{-2}\) and the unit C/(N m\(^{-2}\) s m\(^2\) V\(^{-1}\) s\(^{-1}\)) simplifies to m. \(1 \text{ C} = 1 \text{ A s}\). \(1 \text{ V} = 1 \text{ J A}^{-1} = 1 \text{ N m A}^{-1}\). \( \text{Pa s m}^2 \text{ V}^{-1} \text{ s}^{-1} = (\text{N s m}^{-2}) \text{ m}^2 (\text{N m A}^{-1})^{-1} \text{ s}^{-1} = \text{N s} (\text{N m A}^{-1})^{-1} \text{ s}^{-1} = \text{N s} (\text{N m A}^{-1})^{-1} \text{ s}^{-1} = \text{N s} \frac{\text{A}}{\text{N m}} \text{ s}^{-1} = \frac{\text{A s}}{\text{m}} \). So the unit is \( \frac{\text{C}}{\frac{\text{A s}}{\text{m}}} = \frac{\text{C m}}{\text{A s}} \). Since 1 C = 1 A s, this simplifies to m. The units work out.
Let's continue the numerical calculation:
\[ r = \frac{3.2 \times 10^{-19}}{48\pi \times 10^{-11}} \text{ m} \] \[ r = \frac{3.2}{48\pi} \times 10^{-19 - (-11)} \text{ m} \] \[ r = \frac{3.2}{48\pi} \times 10^{-8} \text{ m} \]Using \( \pi \approx 3.14159 \):
\[ 48\pi \approx 48 \times 3.14159 \approx 150.79632 \] \[ r \approx \frac{3.2}{150.79632} \times 10^{-8} \text{ m} \] \[ r \approx 0.021219 \times 10^{-8} \text{ m} \] \[ r \approx 2.1219 \times 10^{-10} \text{ m} \]We need to express the radius in picometers (pm). Recall that \( 1 \) m \( = 10^{12} \) pm.
\[ r \approx 2.1219 \times 10^{-10} \times 10^{12} \text{ pm} \] \[ r \approx 2.1219 \times 10^{2} \text{ pm} \] \[ r \approx 212.19 \text{ pm} \]Rounding to the nearest whole number based on the options, the effective radius is approximately 212 pm.
Comparing with Options
Let's compare our calculated effective radius with the given options:
- 106 pm
- 212 pm
- 424 pm
- 318 pm
Our calculated value of approximately 212 pm matches option 2.
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