2. HCOs can exist in water either undissociated form, H2COs or dissociated forms

Question

2. HCOs can exist in water either undissociated form, H2COs or dissociated forms, HCO and cos-. But their relative concentrations depend on pH. FECO) with pK,-6.35, pK2 = 10.33 (total 15 points) At pH-12, what species is the primary component? (5 points) (a) (b) At what pH, the concentration of H CO, is equal to that of HC0,? (5 points) (c) At what pH, the concentration of HCO; is equal to that of CO,s3? (5 points) In Lab #4, you prepared NaCl (FW 58.5) solution by dissolving 0.2 g of NaCl with a final volume of 200 mL for the titration with unknown AgNO, solution (15 points) (a) What's the molarity (M) of NaCI solution? (5 points) (b) By the titration, you found that 24.2 mL of NaCI solution you prepared is required for 25 mL of unknown AgNO, solution to reach the end point. What's the concentration of Ag+ in AgNOs solution? Use the following reaction for the calculation (5 points) Ag (aq) +Cl(aq)-AgCI(s) Calculate [Ag+] at the equivalence point (Ksp of AgCl(s)?1.8x10io)DONOT co activity of ions (5 points) (c)

Explanation / Answer

1.

(a)

[HCl] = 10-8 M

So, [H+] =10-8 M

pH = -log (10-8)

pH = 8

(b)

CH3COOH       =        H+        +        CH3COO-
IC:           0.03                      0                             0
C:             - x                      + x                         + x
EC:      0.03 – x                    x                             x

So, Ka = [CH3COO-] [H+] / [CH3COOH]

1.80 x 10-5 = (x) (x) / (0.03 – x)

1.80 x 10-5 = x2 / (0.03 – x)

1.80 x 10-5 = x2 / 0.03    (Ka is very small, so, x term in denominator is neglected)

x2 = (1.80 x 10-5) x 0.03

x2 = 5.4 x 10-7

x = 7.35 x 10-4

So, [H+] = x = 7.35 x 10-4

pH = -log [H+] = - log (7.35 x 10-4) = 3.13

(c)

NH4+                      <===>                  H+           +           NH3
IC:           0.03                                                      0                            0
C:             - x                                                      + x                        + x
EC:      0.03 – x                                                    x                             x

So, Ka = [H+] [NH3] / [NH4+]

Kb = 1.75 x 10-5

Ka = Kw / Kb
      = (1.0 x 10-14) / (1.75 x 10-5)
      = 5.71 x 10-10

So,

5.71 x 10-10 = (x) (x) / (0.03 – x)

5.71 x 10-10 = x2 / (0.03 – x)

5.71 x 10-10 = x2 / 0.03    (Ka is very small, so, x term in denominator is neglected)

x2 = (5.71 x 10-10) x 0.03

x2 = 1.71 x 10-11

x = 4.14 x 10-6

So, [H+] = x = 4.14 x 10-6

pH = -log [H+] = - log (4.14 x 10-6) = 5.38

(d)

CH3COO-     +    H2O    <===>      CH3COOH        +        OH-
IC:           0.03                                                      0                            0
C:             - x                                                      + x                        + x
EC:      0.03 – x                                                    x                             x

So, Kb = [CH3COOH] [OH-] / [CH3COO-]

Ka = 1.80 x 10-5

Kb = Kw / Ka
      = (1.0 x 10-14) / (1.80 x 10-5)
      = 5.56 x 10-10

So,

5.56 x 10-10 = (x) (x) / (0.03 – x)

5.56 x 10-10 = x2 / (0.03 – x)

5.56 x 10-10 = x2 / 0.03    (Ka is very small, so, x term in denominator is neglected)

x2 = (5.56 x 10-10) x 0.03

x2 = 1.67 x 10-11

x = 4.09 x 10-6

So, [OH-] = x = 4.09 x 10-6

pOH = -log [OH-] = - log (4.09 x 10-6) = 5.39

pH = 14 – pOH
     = 14 – 5.39
     = 8.61

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