Chemistry, 29.02.2020 00:27 haleyrene2663

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)

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Answer from: danam04

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

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

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

Answer from: Quest

the number of valence electrons for each molecule or ion is shown beneath the structure. remember hydrogen will not have more than two electrons. this hydrogen is part of a covalent bond (sharing two electrons). a double bond here would cause hydrogen to share four electrons with phosphorus.

Answer from: Quest

2h2o2--> 2h2o + o2

this reaction is of the spontaneous decomposition of hydrogen peroxide down into water and oxygen.

add 2 molecules of hydrogen peroxide and 2 molecules of water. since oxygen is naturally diatomic, the total number of atoms of each element is now the same on both sides of the equation so it is balanced.

this may not be entirely correct but let me know.