1. A solution of nitrous acid, HNO2, was prepared by dissolving 1.93 g of HNO2 i

Question

1. A solution of nitrous acid, HNO2, was prepared by dissolving 1.93 g of HNO2 in 500.0 mL of solution. If the equilibrium concentration of H+ in this solution is 7.85 x 10-3 M, what is the equilibrium constant for the dissociation reaction: HNO2 H+ + NO2-

2. For the following dissociation of acetic acid, K = 1.76 x 10-5: CH3COOH CH3COO- + H+
Sodium acetate completely dissociates in solution according to the following reaction:NaCH3COO Na+ + CH3COO-

A solution was prepared which contains CH3COOH at a molar analytical concentration of 0.438 M and NaCH3COO at a molar analytical concentration of 0.362 M.What is the equilibrium concentration of H+?

3. For the following dissociation of acetic acid, K = 1.76 x 10-5: CH3COOH CH3COO- + H+

Sodium acetate completely dissociates in solution according to the following reaction:NaCH3COO Na+ + CH3COO-

A solution was prepared which contains NaCH3COO at a molar analytical concentration of 0.189 M and an equilibrium concentration of H+ at 7.34 x 10-5 M. What is the equilibrium concentration of acetic acid?

4. For the following dissociation of acetic acid, K = 1.76 x 10-5: CH3COOH CH3COO- + H+

Sodium acetate completely dissociates in solution according to the following reaction:NaCH3COO Na+ + CH3COO-

If 701.6 mL of an acetic acid solution with a molar analytical concentration of 7.95 M is added to 454.8 mL of a sodium acetate solution with a molar analytical concentration of 0.763 M, what is the equilibrium concentration of H+ in the resulting solution?

5. For the following dissociation of acetic acid, K = 1.76 x 10-5: CH3COOH CH3COO- + H+

Sodium acetate completely dissociates in solution according to the following reaction:NaCH3COO Na+ + CH3COO-

What mass (mg) of sodium acetate must be added to 559.8 mL of an acetic acid solution with a molar analytical concentration of 0.393 M to obtain an equilibrium concentration of hydrogen 4.27 x 10-4 M?

Assume the addition of sodium acetate does not change the volume of the solution.

Explanation / Answer

Given :

Mass of HNO2 = 1.93 g

Volume of solution = 500.0 mL

Equilibrium [H+] = 7.85 E-3 M

Equilibrium constant = Unknown

Solution:

Calculation of Molarity of HNO2

[HNO2] = # mol HNO2 / volume in L

Mol HNO2 = Mass in g / Molar mass

= 1.93 g / 47.0129 g per mol

= 0.04105

[HNO2]= 0.04105 mol / 0.500 L

= 0.0821 M

Lets write the reaction and set up ICE chart.

Reaction

HNO2       H+   +    NO2 -

I               0.0821         0                      0

C             -x                     +x                +x

E          (0.0821-x)               x              x

x = [H+] = 7.85 E-3 M

we use this value to get equilibrium concentration of each species.

[HNO2]= 0.0821 – 7.85 E-3 = 0.0743 M

[H+] = 7.85 E-3 M

[NO2]= x = 7.85 E-3

Calculation of equilibrium constant

K = [H+] [ NO2-]/ [HNO2]

   = (7.85E-3 M ) 2 / ( 0.0743 )

= 8.3 E-4

Q. 2

In this problem we find the buffer system

For buffer we use Henderson-Hasselbalch equation

pH = pka + log ([Conj base]/[Acid])

Here acid is acetic acid so pka of acetic acid

= - log ka

Ka of acetic acid = 1.76 E-5

pka = - log ka

= -log ( 1.76 E-5 )

= 4.75

Here [NaAcetate ]= [Acetate ion]

Acetate ion is conjugate base of acetic acid

Lets plug all the concentration in above equation to get pH

pH = 4.75 + log ( [ 0.362 ]/[0.438] )

pH = 4.67

We have to find concentration of H+ at equilibrium

We use following equation

pH = -log [H+]

[H+] = Antilog ( -pH )

= Antilog ( - 4.67)

= 2.13 E-5 M

[H+] = 2.13 E-5 M

3.

In this problem we use same equation since it is buffer system

Calculation of pH

pH = -log [H+]

= -log ( 7.34 E-5 M)

=4.13

Lets plug all the value

4.13 = 4.75 + log ( [ 0.189 M / [ acetic acid ] )

-0.62 = log ( [ 0.189 M / [ acetic acid ] )

Lets take antilog of both side

0.2398 = ( [ 0.189 M / [ acetic acid ] )

[Acetic acid ] = 0.788 M

Q. 4 )

Here we find concentration of each acetic acid and acetate ion.

[CH3COOH] = mol / Volume in L

Volume is total volume and moles that we get by using concentration and volume in L

Moles of acetic acid = Volume in L * Concentration

= 0.7016 L * 7.95 mol /L

= 5.58 mol

[CH3COOH ] = 5.58 mol / ( 0.7016 L + 0.4548 )L

= 4.82 M

Calculation of concentration of sodium acetate

Number of moles of sodium acetate

= 0.4548 L * 0.763 mol/ L

=0.3470 M

pH = 4.75 + log ( [ 0.3470 ] / [ 4.82 ] )

= 3.60

pH = 3.60

[H+] = Antilog ( - pH )

= Antilog ( - 3.60 )

=2.47 E-4 M

Q. 5

We calculate concentration of sodium acetate by using Henderson-Hasellbalch equation

Calculation of H+

[H+] = - log (4.27 E-4 M)

= 3.67

3.67 = 4.75 + log ( [ sodium acetate ] / 0.393 )

-1.38 = log ( [ sodium acetate ] / 0.393 )

Lets take antilog of both sides

0.0416 = ( [ sodium acetate ] / 0.393 )

[Sodium acetate ]= 0.01637 M

We use volume to get moles of sodium acetate

Mol of sodium acetate = 0.5598 L * 0.01637 M

= 0.00916 mol

Calculation of mass of sodium acetate

= Mole * molar mass

= 0.00916 mol * 82.0343 g per mol

= 0.752 g sodium acetate

Mass of sodium acetate needed = 0.752 g

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