For 520.0 mL of a buffer solution that is 0.135 M in HC2H3O2 and 0.120 M in NaC2

Question

For 520.0 mL of a buffer solution that is 0.135 M in HC2H3O2 and 0.120 M in NaC2H3O2, calculate the initial pH and the final pH after adding 0.020 molof HCl.

Express your answers using two decimal places separated by a comma.

pHinital, pHfinal=

Part B:

For 520.0 mL of a buffer solution that is 0.145 M in CH3CH2NH2 and 0.135 M in CH3CH2NH3Cl, calculate the initial pH and the final pH after adding 0.020 mol of HCl.

Express your answers using two decimal places separated by a comma.

pHinital, pHfinal=

Part B:

For 520.0 mL of a buffer solution that is 0.145 M in CH3CH2NH2 and 0.135 M in CH3CH2NH3Cl, calculate the initial pH and the final pH after adding 0.020 mol of HCl.

Express your answers using two decimal places separated by a comma.

pHinitial, pHfinal =

Explanation / Answer

Q1.

a) Initial pH of Buffer

Since this is a buffer; consider using the Henderson Hasselbach equation, it helps to model buffers

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

the pKa value is the power funciton of the Ka; equilibrium constant for the acid. It is given as

pKa = -log(Ka)

Ka value for acetic acid (HC2H3O2) = 1.8*10^-5

pKa = -log(1.8*10^-5) = 4.75

[conjugate] = concentration of the conjugate, that is, the "acetate" ion or NaC2H3O2

[NaC2H3O2] = 0.12

[acid] = concentration of the acid, that is, acetic acid HC2H3O2

[HC2H3O2] = 0.135

then

substitute in the equation

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

pH = 4.75 + log(0.135/0.120) = 4.8

the value is pretty near to that of the pKa, meaning it is in the buffer region, as expected.

b)

Calculate pH After adding 0.02 mol of HCl

the solution will get more acidic, so expect lower pH than previous one

calcualte [HCl] = mol/V

V = 520 mL = 0.52 L

[HCl] = 0.02 / 0.520 = 0.03846 M

then

after addition:

HCl --> H+ + Cl-

NaC2H3O2 + H+ --> HC2H3O2 + Na+

then, we are increasing C2H3O2 concentration; and lowering NaC2H3O2

this can be modeled as

[NaC2H3O2] = 0.12 - 0.03846 = 0.08154 M

[HC2H3O2] = 0.135 + 0.03846 = 0.17346 M

with these new concentrations, we may now substitute in the pH equation

remember that the acid has increased and the acetate has lowered its concentration

therefore

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

pH = 4.75 + log(0.08154/0.17346) = 4.4221713

pH = 4.4221713

which is lower as the previous example, as expected

Q2

This is pretty similar to the previous example

but now we deal with a weak BASE. Or a Basic buffer

this can be modelled prety similar to the previous Acidic buffer

pOH = pKb + log([conjugate]/[base])

CH3CH2NH2 is ethylamine, fint its Kb or pKb value in datatables

Kb = 4.3*10^-4

calculate pKb as follows:

pKb = -log(kb) = -log(4.3*10^-4) = 3.366531

now...

pOH = pKb + log([conjugate]/[base])

identify the base; which is of course CH3CH2NH2

identify the conjugate... should be the ionic state of the ethylamine. that is CH3CH2NH3+ (which comes from CH3CH2NH3Cl)

then

initially:

pOH = 3.366531 + log(0.135/ 0.145)

pOH = 3.33549

we need pH so; from the equality

14 = pH + pOH

pH = 14-pOH = 14-3.33549

pH = 10.66451

b.

after addition of 0.02 mol of HCl

we will have a neutralization, since acid + base react to form water and salt

HCl -> H+ + Cl-

H+ + OH- = H2O

which comes from:

CH3CH2NH2 + H2O <--> CH3CH2NH3+ + OH-

CH3CH2NH3+ + OH- + H+ = CH3CH2NH3+ + H2O

note that CH3CH2NH3+ is the conjguate, so we are forming more conjguate, and we are getting rid of the base CH3CH2NH2

then

[HCl] = mol/V

[HCl] = 0.02/0.520 = 0.0384 M

then

[conjugate] being formed = 0.135 + 0.0384 = 0.1734

[base] remaining after reaciton = 0.145 - 0.0384 = 0.1066

as expected, the base concentration decreases. The conjguate increases

then

the pH must decrease, that is, become more acidic

then

pOH = 3.366531 + log(0.1734/0.1066 )= 3.577822

but we need pH so

pH = 14-pOH = 14-3.577822

pH = 10.422178; which is stil basic, but less basic that the previous example

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