Question

In: Chemistry

1. A 29.3 mL sample of 0.369 M ethylamine, C2H5NH2, is titrated with 0.205 M hydrochloric...

1. A 29.3 mL sample of 0.369 M ethylamine, C2H5NH2, is titrated with 0.205 M hydrochloric acid.
After adding 74.4 mL of hydrochloric acid, the pH is:

2. A 26.4 mL sample of 0.296 M diethylamine, (C2H5)2NH, is titrated with 0.206 M hydrobromic acid.
After adding 15.2 mL of hydrobromic acid, the pH is:

3. Calculate the pH and the equilibrium concentrations of HCO3- and CO32- in a 0.0510 M carbonic acid solution, H2CO3 (aq).
For H2CO3, Ka1 = 4.2×10-7 and Ka2 = 4.8×10-11

pH=
[HCO3-]=
[CO32-]=

4. Calculate the pH and the equilibrium concentration of CO32- in a 5.13×10-2 M carbonic acid solution, H2CO3 (aq).
For H2CO3, Ka1 = 4.2×10-7 and Ka2 = 4.8×10-11

pH=
[CO32-]=

Solutions

Expert Solution

1) millimoles of ethylamine base = 29.3 x 0.369 = 10.81

millimoles of HCl = 74.4 x 0.205 = 15.25

acid dominent : so acid concentration remains = 15.25 - 10.81 / (29.3 + 74.4) = 0.043 M

pH = -log [H+] = -log (0.043) = 1.37

pH = 1.37

2)

base millimoles = 26.4 x 0.296 = 7.81

acid millimoles = 0.206 x 15.2 = 3.13

B +    H+ ----------------> salt

7.81    3.13                        0

4.68     0                             3.13

it is buffer

pOH = pKb + log [salt / base]

pOH = 3.16 + log (3.13 / 4.68)

pOH = 2.98

pH + pOH = 14

pH = 11.02

3)

answers :

[HCO3-] = 1.46 x 10^-4 M

pH = 3.83

[CO3-2] = 4.8 x 10^-11 M


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