Assignment Answer Page

1] First calculate mass of solute in a Kg of solvent:

Now consider that dominator in molality can be converted to a solution volume:

2]     x-bar (average) = 0.116        s = 0.0033

        from table 4-2 95% C.L. and df = 3, t = 3.182

        μ = 0.116 ± (3.182*0.0033)/Ö4 = 0.0116 ± 0.0052

        Now consider that 0.123 – 0.116 = 0.007

        We see that 0.007 > 0.0052, so to the 95% C.L. we can assume a determinate error.

3]     Plot signal vs. [Cd²⁺]_spike.

    Conc of added standard = (10e-6 L * 1.51e-3M)/0.100-L = 1.51e-7 M 

Slope = (6.77 – 4.31 mA) / (1.51e-7 M - 0) = 1.63e7 uA / M

y-int = 4.31 uA

line:      y =  (1.63e7 uA / M) x + 4.31 uA

x-int:     0 =  (1.63e7 uA / M) x + 4.31 uA

            x = - 2.65e-7 M

conc in 100-gm sample     2.65e-7 M * 0.100-L * 112.411 g/mol * 10e9/100-gm = 29.8 ppb

4] CaCO₃ K_sp = 4.5e-9 = [Ca²⁺][CO₃^2-]

 MBE:   [Ca²⁺] = [CO₃^2-] + [HCO₃^-] + [H₂CO₃]

 Hydrolysis of CO₃^2-

 CO₃^2- + H₂O = HCO₃^- + OH^-              K_b1 = K_w/K_a1 = 1.01e-14/4.69e-11 = 2.15e-4

 HCO₃^- + H₂O = H₂CO₃ + OH^-            K_b2 = K_w/K_a2 = 1.01e-14/4.45e-7 = 2.27e-8

 Also for pH = 4.2, pOH = 9.8 and [OH^-] = 1.6e-10

 K_b1 = [HCO₃^-][OH^-]/[CO₃^2-]                            solve for [HCO₃^-] for sub into MBE

[HCO₃^-] = K_b1[CO₃^2-]/[OH^-] 

K_b2 = [H₂CO₃][OH^-]/[HCO₃^-]                          solve for [H₂CO₃] for sub into MBE

[H₂CO₃] = K_b2[HCO₃^-]/[OH^-]              use [HCO₃^-] = K_b1[CO₃^2-]/[OH^-]

[H₂CO₃] = K_b2 K_b1[CO₃^2-]/[OH^-]²

Now we have

[Ca²⁺] = [CO₃^2-] + K_b1[CO₃^2-]/[OH^-] + K_b2 K_b1[CO₃^2-]/[OH^-]²

At pH 4.2

[Ca²⁺] = [CO₃^2-] + 1.3e6[CO₃^2-] + 2e8[CO₃^2-]

[Ca²⁺] = 2e8[CO₃^2-]

And we have K_sp = 4.5e-9 = [Ca²⁺][CO₃^2-]

[CO₃^2-] = 4.5e-9/[Ca²⁺]

[Ca²⁺] = 2e8*4.5e-9/[Ca²⁺]

[Ca²⁺] = 0.9 M

Note that pH 7.0, [OH^-] = 1.0e-7

[Ca²⁺] = [CO₃^2-] + K_b1[CO₃^2-]/[OH^-] + K_b2 K_b1[CO₃^2-]/[OH^-]²

            [Ca²⁺] = 2.6e3*[CO₃^2-]

            [Ca²⁺] = 2.6e3*4.5e-9/[Ca²⁺]

            [Ca²⁺] = 3e-3 M

5]    Let x = grams of NaBr and y = grams of NaCl

Eqn 1: x + y = 0.2500 g

x-g NaBr * (mol NaBr/102.89 g) *  (187.80 g AgBr /mol) = 1.8253 x

y-g NaCl * (mol NaCl/58.22 g) * (143.35 g AgCl/mol) = 2.4622 y

Eqn 2: 1.8253 x + 2.4622 y = 0.5348 g

x = 0.2500 - y

            1.8253*(0.2500 – y) + 2.4622 y = 0.5348

            0.4563 - 1.8253 y + 2.4622 y = 0.5348

            0.6369 y = 0.0785

            y = 0.123 g                  &         x = 0.127 g

6] Calculate pCl for the following titration:

            a) 20.00 mL of 0.0600 M Hg₂(NO₃)₂ is added to 4.00 mL of 0.150 M NaCl

            b) 20.00 mL of 0.0600 M Hg₂(NO₃)₂ is added to 16.00 mL of 0.150 M NaCl

            c) 20.00 mL of 0.0600 M Hg₂(NO₃)₂ is added to 32.00 mL of 0.150 M NaCl

Hg₂Cl₂  K_sp = 1.3e-18

a)                  20.00 mL*0.0600 M = 1.20 mmol Hg₂²⁺

4.00 mL*0.150 M = 0.600 mmol Cl^-

                        Find limiting reagent:

            0.600 mmol Cl^- * (1 mol Hg₂²⁺ / 2 Cl^-) = 0.300 mmol Hg₂²⁺

            1.20 mmol Hg₂²⁺ * (2 mol Cl^-/ 1 Hg₂²⁺) =  2.40 mmol Cl^-

            Therefore Cl^- is the L.R.

            Hg₂²⁺    +          2Cl^-      =          Hg₂Cl₂(s)

            1.20 mmol        0.600 mmol      --

            -1/2*0.600       -6.00                --

            0.900 mmol      0                      --

            Excess [Hg₂²⁺] = (1.20 – 0.300) mmol / 24.00 mL = 0.0375 M

            K_sp = 1.3e-18 = [Hg₂²⁺] [Cl^-]²

            [Cl^-] = (1.3e-18 / 0.0375)^1/2 = 5.9e-9

            pCl = 8.23

b)                  This is the equivalence point.

20.00 mL*0.0600 M = 1.20 mmol Hg₂²⁺

16.00 mL*0.150 M = 2.40 mmol Cl 

Hg₂²⁺          +          2Cl^- à             Hg₂Cl₂ 

x                            2x

1.3e-18 = [Hg₂²⁺] [Cl^-]² = x(2x)² = 4x³

x = 6.9e-7

            NOTE that the solubility of Cl- is 2x and NOT x. The solubility of Hg₂²⁺ is x.

[Cl-] = 2x = 1.4e-6

            pCl = 5.86

c)         20.00 mL*0.0600 M = 1.20 mmol Hg₂²⁺

32.00 mL*0.150 M = 4.80 mmol Cl^-

All Hg₂²⁺ is consumed leaving excess Cl-

Hg₂²⁺    +          2Cl^-      =          Hg₂Cl₂(s)

            1.20 mmol        4.80                 --

            -1.20                -2*1.20            --

            0.0 mmol          2.40 mmol        --

Excess [Cl-] = [4.80 mmol – (2*1.20 mmol)] / 52.00 mL = 4.62e-2

pCl = 1.336

7] Calculate the pH when the following volumes of 0.100 M KOH is added 50.00 mL of 0.150 M of a weak acid HA whose K_a is 8.0e-5

a)      50.00 mL

b)      75.00 mL

c)      100.00 mL

a] 50.00 mL                 f = (50.00 mL*0.100 M KOH)/(50.00 mL*0.150) = 0.667 before eq. pt.

HA       +          OH^-      à        A^-         +          H₂O

7.50 mmol        5.00 mmol        0                      --

-5.00                -5.00                +5.00               --

2.50                 0                      +5.00

Buffer region

Ka = [H⁺][A^-]/[HA]                 8.0e-5 =  [H⁺][5.00]/[2.50]                  [H⁺] = 4.0e-5

pH = 4.40

b] 75.00 mL                f = (75.00 mL*0.100 M KOH)/(7.50 mmol) = 1 at the eq. pt.

HA       +          OH^-      à        A^-         +          H₂O

7.50 mmol        7.50 mmol        0                      --

-7.50                -7.50                +7.50               --

0                      0                      +7.50

Must realize that A^- hydrolyzes.

            A^-         +          H₂O     à        HA       +          OH^-

            7.50/125          --                     0                      0

            -x                     --                     +x                    +x

            7.50/125-x                   --           x                      x

K_b = K_w/K_a = 1.01e-14/8.0e-5 = 1.26e-10 = x²/0.0600-x ≈ x²/0.0600

x = [OH^-] = 2.75e-5                 pOH = 5.56                 pH = 8.44

c] 100.00 mL               f = (100.00 mL*0.100 M KOH)/(7.50 mmol) = 1.33 past eq. pt.

HA       +          OH^-      à        A^-         +          H₂O

7.50 mmol        10.0 mmol        0                      --

-7.50                -7.50                +7.50               --

0                      2.50                 +7.50

[OH^-] = 2.50 mmol/150.00 mL = 1.66e-2 M

pOH = 1.778               pH = 12.222

8] Calculate pCa for a titration in which 50.00 mL of 2.00e-2 M Ca(NO₃)₂ is titrated with 0.100 M EDTA at pH 9.00 solution.

a)      0.00 mL

b)      5.00 mL

c)      10.00 mL

d)      20.00 mL

At 0.00 mL      [Ca²⁺] = 2.00e-2 M    

 pCa = 1.699            K_f’ = a_Y4-K_f = 5.4e-2*4.9e10 = 2.6e9

At 5.00 mL

50.00 mL*2.00e-2 M = 1.00 mmol Ca²⁺

5.00 mL*0.100 M = 0.500 mmol EDTA

Ca²⁺     +          EDTA              =          CaY^2-

1.00 mmol        0.500                           0

-0.500              -0.500                          +0.500

0.500               0                                  0.500 mmol

[Ca²⁺] = 0.500 mmol/55.00 mL = 9.09e-3

pCa = 2.041

at 10.00 mL

10.00 mL*0.100 M = 1.00 mmol EDTA

Ca²⁺     +          EDTA              =          CaY^2-

1.00 mmol        1.00                             0

-1.00                -1.00                            +1.00

0                      0                                  1.00 mmol

[CaY^2-] = 1.00 mmol/60.00 mL = 1.67e-2 M

Ca²⁺     +          EDTA              =          CaY^2-

0                      0                                  1.67e-2

+x                    +x                                -x

x                      x                                  1.67e-2 – x

K_f’ = 2.6e9 = [1.67e-2 - x]/x²

x = 2.49e-6

pCa = 5.603

at 20.00 mL

20.00 mL*0.100 M = 2.00 mmol EDTA

Ca²⁺     +          EDTA              =          CaY^2-

1.00 mmol        2.00                             0

-1.00                -1.00                            +1.00

0                      1.00                             1.00 mmol

K_f’ = 2.6e9 = [CaY^2-]/[Ca²⁺][EDTA]

            [CaY^2-] = 1.00 mmol/70.00 mL = 1.429e-2 M

            [EDTA] = 1.00 mmol/70.00 mL = 1.429e-2

2.6e9 = 1.429e-2/[Ca²⁺]1.429e-2

[Ca²⁺] = 3.8e-10

pCa = 9.415

9]

At 0.00 mL

[Zn²⁺] = a_Zn2+C_Zn2+ = 1.79e-5*1.00e-3 = 1.79e-8

pZn = 7.747

2.50 mL

Added EDTA = 2.50 mL*2.0e-3 M = 0.0050 mmol                before eq. pt.

                        C_Zn+     +          EDTA              =          ZnY^2-

                        0.0100             0.0050                         0          mmol

                        -0.0050            -0.0050                        +0.0050

                        0.0050             0                                  0.0050

[Zn²⁺] = a_Zn2+C_Zn2+ = 1.79e-5*(0.0050/12.50) M = 7.2e-9

pZn = 8.15

5.00 mL

Added EDTA = 5.00 mL*2.0e-3 M = 0.0100 mmol                eq. pt.

                        C_Zn+     +          EDTA              =          ZnY^2-

                        0.0100             0.010                           0          mmol

                        -0.010              -0.010                          +0.010

                        0                      0                                  0.010

[ZnY^2-] = 0.010 mmol/15.00 mL = 6.7e-4

                        C_Zn+     +          EDTA              =          ZnY^2-

                        0                      0                                  6.7e-4 M

                        +x                    +x                                -x

                        x                      x                                  6.7e-4-x

K_f” = 4.8e11 = [ZnY^2-] /C_Zn2+[EDTA] = 6.7e-4/x²

x = 3.7e-8 M = C_Zn+

[Zn²⁺] = a_Zn2+C_Zn2+ = 1.79e-5*3.7e-8 M = 6.7e-13

pZn = 12.17

10.00 mL

Added EDTA = 10.00 mL*2.0e-3 M = 0.020 mmol                past eq. pt.

                        C_Zn+     +          EDTA              =          ZnY^2-

                        0.0100             0.020                           0          mmol

                        -0.010              -0.010                          +0.010

                        0                      0.010                           0.010

K_f” = 4.8e11 = [ZnY^2-] /C_Zn2+[EDTA] = 0.010/ C_Zn2+[0.010]

C_Zn2+ = 2.1e-12

[Zn²⁺] = a_Zn2+C_Zn2+ = 1.79e-5*2.1e-12 = 3.7e-17

pZn = 16.43