Question

For a solution comprised of PbSO4 (0.025 mol/kg) and MgCl2 (0.01 mol/kg) determine (19 pts) (a) the ionic strength (5 pts) (b) activity coefficients of each of these ions in solution (6 pts)

Answer 1

Answer:

For a: The ionic strength of lead sulfate solution is 0.1 mol/kg and for magnesium chloride is 0.03 mol/kg

For b: The activity coefficient of $Pb^{2+}\text{ and }SO_4^{2-}$ ions in lead sulfate solution is 0.352 and 0.324 respectively and the activity coefficient of $Mg^{2+}\text{ and }Cl^{-}$ ions in magnesium chlorirde solution is 0.500 and 0.843 respectively

Explanation:

• For a:

To calculate the ionic strength, we use the equation:

$I=\frac{1}{2}\sum_{i=1}^n(C_iZ_i^2)$

where,

$C_i$ = concentration of i-th ions.

$Z_i$ = charge of i-th ions.

Taking $PbSO_4$ solution:

The chemical equation for the ionization of lead sulfate follows:

$PbSO_4(aq.)\rightarrow Pb^{2+}(aq.)+SO_4^{2-}(aq.)$

For the ions:

$C_{SO_4^{2-}}=0.025mol/kg\\C_{Pb^{2+}}=0.025mol/kg\\Z_{SO_4^{2-}}=-2\\Z_{Pb^{2+}}=+2$

Putting values in above equation, we get:

$I=\frac{1}{2}[(0.025\times (+2)^2)+(0.025\times (-2)^2)]\\\\I=0.1mol/kg$

Taking $MgCl_2$ solution:

The chemical equation for the ionization of lead sulfate follows:

$MgCl_2(aq.)\rightarrow Mg^{2+}(aq.)+2Cl^{-}(aq.)$

For the ions:

$C_{Cl^{-}}=0.02mol/kg\\C_{Mg^{2+}}=0.01mol/kg\\Z_{Cl^{-}}=-1\\Z_{Mg^{2+}}=+2$

Putting values in above equation, we get:

$I=\frac{1}{2}[(0.01\times (+2)^2)+(0.02\times (-1)^2)]\\\\I=0.03mol/kg$

Hence, the ionic strength of lead sulfate solution is 0.1 mol/kg and for magnesium chloride is 0.03 mol/kg

• For b:

To calculate the activity coefficient, we use the equation:

$-\log \gamma_i=\frac{0.509Z_i^2\sqrt{I}}{1+3.3(\alpha_i \sqrt{I})}$

where,

$Z_i$ = charge on i-th specie

I = ionic strength

$\alpha$ = effective hydrated radius of i-th specie (in nm)

• Taking $PbSO_4$ solution:

For $Pb^{2+}$ ions:

We know that:

$\alpha_{Pb^{2+}}=0.401nm\\I=0.1mol/kg\\Z_{Pb^{2+}}=+2$

Putting values in above equation, we get:

$-\log \gamma_{Pb^{2+}}=\frac{0.509\times (+2)^2\sqrt{0.1}}{1+3.3(0.401\sqrt{0.1})}\\\\-\log \gamma_{Pb^{2+}}=0.454\\\\\gamma_{Pb^{2+}}=10^{-0.454}=0.352$

For $SO_4^{2-}$ ions:

We know that:

$\alpha_{SO_4^{2-}}=0.300nm\\I=0.1mol/kg\\Z_{SO_4^{2-}}=-2$

Putting values in above equation, we get:

$-\log \gamma_{SO_4^{2-}}=\frac{0.509\times (-2)^2\sqrt{0.1}}{1+3.3(0.300\sqrt{0.1})}\\\\-\log\gamma_{SO_4^{2-}}=0.490\\\\\gamma_{SO_4^{2-}}=10^{-0.490}=0.324$

• Taking $MgCl_2$ solution:

For $Mg^{2+}$ ions:

We know that:

$\alpha_{Mg^{2+}}=0.300nm\\I=0.03mol/kg\\Z_{Mg^{2+}}=+2$

Putting values in above equation, we get:

$-\log \gamma_{Mg^{2+}}=\frac{0.509\times (+2)^2\sqrt{0.03}}{1+3.3(0.300\sqrt{0.03})}\\\\-\log\gamma_{Mg^{2+}}=0.301\\\\\gamma_{Mg^{2+}}=10^{-0.301}=0.500$

For $Cl^{-}$ ions:

We know that:

$\alpha_{Cl^{-}}=0.330nm\\I=0.03mol/kg\\Z_{Cl^{-}}=-1$

Putting values in above equation, we get:

$-\log \gamma_{Cl^{-}}=\frac{0.509\times (-1)^2\sqrt{0.03}}{1+3.3(0.330\sqrt{0.03})}\\\\-\log\gamma_{Cl^{-}}=0.0742\\\\\gamma_{Cl^{-}}=10^{-0.0742}=0.843$

Hence, the activity coefficient of $Pb^{2+}\text{ and }SO_4^{2-}$ ions in lead sulfate solution is 0.352 and 0.324 respectively and the activity coefficient of $Mg^{2+}\text{ and }Cl^{-}$ ions in magnesium chlorirde solution is 0.500 and 0.843 respectively