1) Consider the reaction: HC2H3O2 + H2O > H3O+ + C2H3O2- Choose the pair of subs

Question

1) Consider the reaction:

HC2H3O2 + H2O > H3O+ + C2H3O2-

Choose the pair of substances that are both bases in the reaction.

HC2H3O2 and H3O+

H2O and C2H3O2-

H2O and H3O+

HC2H3O2 and C2H3O2-

all are bases

Show work and explain for the following

2)The base-dissociation constant of ethylamine (C2H5NH2) is 6.4 × 10-4 at 25.0°C. What is the [H+] in a 1.3 × 10-2 M solution of ethylamine?

3)A 100.0 mL sample of 0.10 M NH3 is titrated with 0.10 M HNO3. Determine the pH of the solution after the addition of 50.0 mL of HNO3. The Kb of NH3 is 1.8 × 10-5.

4)What is the pH of a buffer system prepared by dissolving 10.70 grams of NH4Cl and 35.00 mL of 12 M NH3 in enough water to make 1.000 L of solution? K b = 1.80 × 10-5for NH3.

5)A 1.00 L buffer solution is 0.250 M in HF and 0.250 M in LiF. Calculate the pH of the solution after the addition of 0.150 moles of solid LiOH. Assume no volume change upon the addition of base. The Ka for HF is 3.5 × 10-4.

6)What is the molar solubility of magnesium carbonate ( MgCO3 ) in water? The solubility-product constant for MgCO3 is 3.5 x 10-8 at 25°C.

HC2H3O2 and H3O+

H2O and C2H3O2-

H2O and H3O+

HC2H3O2 and C2H3O2-

all are bases

Explanation / Answer

1) the acid is one which donates proton or obtained from a base by accepting proton ( so it wil be conjugated acid)

HC2H3O2 + H2O > H3O+ + C2H3O2-

Here HC2H3O2 loses proton and therefore is an acid

H3O+ is obtained from water by accepting proton , so it conjugate acid of H2O.

Now for base

the base is one which accept proton or obtained from an acid by donation of proton ( so it wil be conjugated base)

HC2H3O2 + H2O > H3O+ + C2H3O2-

Here H2O accept proton and therefore is a base

C2H3O2- is obtained from HC2H3O2 by donation of proton , so it conjugate base of HC2H3O2.

2)           CH3CH2NH2 + H2O <=> CH3CH2NH3+ + OH-
Initial      1.6 x 10^-2                             0                  0
change     -x. . . . . . . . . . . . . . . . . . . .+x. . . . . . . . +x

So Kb = 6.4 x 10^-4 = x^2 / 1.6 x 10^-2 -x

1.0 x 10^-5 - 6.4 x 10^-4 x = x^2

0 = x^2 - 1.0 x 10^-5 + 6.4 x 10^-4 x

x= 0.0028
[OH-]= 0.0028M
pOH = 2.55
pH = 14 - 2.5 = 11.45

pH = -log[H+]

so [H+] = 3.54 X 10^-12 M

3) moles NH3 = 0.10 M X 0.100 L= 0.0100
moles HNO3 = 0.10 x 0.0500 = 0.005

NH3 + H+ --> NH4+
moles NH3 = 0.0100 - 0.005 = 0.005
moles NH4+ = 0.005
total volume = 0.150 L
[NH3]= 0.0050 / 0.150 = 0.033M
[NH4+]= 0.005/ 0.150 = 0.033 M

pOH = 4.7 + log [salt /acid]

pOH = 4.7 + log 0.033 / 0.033 = 4.7

sp pH = 14- 4.7 = 9.3

4) Moles NH4Cl = 10.70 g / 53.49 g/mol =0.2000
Moles NH3 = 0.035 L x 12 =0.4200
[NH4+]= 0.2000 / 1.000 L = 0.2000
[NH3] = 0.4200 / 1.000 = 0.4200
NH3 + H2O <----> NH4+ + OH-
1.80 x 10^-5 = ( 0.2000 +x) (x) / 0.4200-x
x = [OH-] = 3.78 x 10^-5 M
pOH =4.42
pH = 14 - pOH =9.58

5) we know from Hendersen Hassalbalch equation

pH = pKa + log [salt] /[acid]

if we will add 0.150 moles of base so the concentration will be 0.150 M

pH = 3.45 + log (0.250 +0.150 )/ (0.250 -0.150) = 3.45 + log 4 = 3.45 +0.602 = 4.052

6) We know that

Ksp = 3.5 X 10^-8 = [Mg+2][CO3-2]

Let solubility = x

So 3.5 X 10^-8 = x^2

x = 1.87 X 10^-4

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