1. A 29.3 mL sample of 0.369 M ethylamine , C 2 H 5 NH 2 , is titrated with 0.20
ID: 969636 • Letter: 1
Question
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-]=
Explanation / Answer
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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