a.)Consider a buffer solution that is 0.50 M in NH3and 0.20 M in NH4Cl. For ammo

Question

a.)Consider a buffer solution that is 0.50 M in NH3and 0.20 M in NH4Cl. For ammonia, pKb=4.75.

Calculate the pH of 1.0 L of the solution, upon addition of 49.00 mL of 1.0 MHCl.

b.)A 1.0-

L buffer solution contains 0.100 molHC2H3O2 and 0.100 molNaC2H3O2. The value of Ka for HC2H3O2 is 1.8×105.

Calculate the pH of the solution upon addition of 19.1 mL of 1.00 MHCl to the original buffer.

C.If you have 500

mL of a 0.10 M solution of the acid, what mass of the corresponding sodium salt of the conjugate base do you need to make the buffer with a pHof 7.68 (assuming no change in volume)?

Express the mass to two significant figures and include the appropriate units.

Explanation / Answer

In a buffer system both species react until an equilibrium has been established.

a.) In the case of NH3/NH4Cl system, the acid dissociation equation at equilibrium is given as:

Kb= {[NH3][H3O+]}/{[NH4+]}

where Kb is the dissociation constant.

When HCl is added to this system, the NH3 will react inorder to maintain this equilibrium. This results in a decrease in NH3 concentration and an increase in the NH4+ concentration thus resulting in a decrease in the pH of the system.

pKa+pKb=14

Therefore pKa=14-4.75 = 9.25 (this is the dissociation constant for NH4+)

Number of moles of HCl = (1M x 49.00 mL)/1000 = 0.049 moles

Number of moles of NH3 = (0.5M x 1000ml)/1000 = 0.5 moles

Number of moles NH4+ = (0.2M x 1000ml)/1000 = 0.2 moles

Using the Henderson-Hasselbalch equation:

pH = pKa + log ([base]/[acid])

therefore: pH = 9.25 + log ((0.5 mol. NH3-0.049 mol. HCl)/(0.5mol NH4+ +).049 mol NH4+))

pH = 9.16

b.) Similar as above however the NaC2H3O2 will react with the HCl

pKa = -log(Ka)

pKa = -log (1.8 x 10-5) = 4.74

Number of moles NaC2H3O2 = 0.1 moles

Number of moles HCl = (19.1mL x 1M)/1000 = 0.0191 moles

pH = 4.74 + log((0.1moles NaC2H3O2 - 0.0191 moles HCl)/(0.1 mol HC2H3O2 +0.0191 moles HC2H3O2))

pH = 4.57

c.) Using the same equation

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

7.68= 4.74 +log([salt]/[acid])

2.94 =log([salt]/[acid])

10-2.94 =[salt]/[acid]

0.001128 = [salt]/[acid]

[salt] = 0.001128 x 0.1M = 1.148 x 10-4M

vol. of solution 500 mL

Number of moles salt = (1.148 x 10-4M x 500mL)/1000 = 5.7408 x 10-5 moles

Mr sodium acetate = 82.0343 g/mol

mass of sodium acetate = 5.7408 x 10-5 moles x 82.0343 g/mol = 4.7 x10-3g

Note: conjugate salt is sodium acetate

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