C. pH of Buffer Solutions In the unshaded portions ofthe following table, pply t

Question

C. pH of Buffer Solutions In the unshaded portions ofthe following table, pply the calculated molarities of the indicated species in the solution mixtures. Assume the solutions are at 25 C. Note that in the rows labeled Before Reaction, you must supply the ate concentrations after dilution occurs in the reaction mixture, but before any reaction takes place. Calculated Molarities in Mixtares [Haj 0,0STM C2a Buffer HCI Before Reaction O.u8M 0.4q M C2b Buffer NaOH Before Reaction 6 M 0.49m At Equilibrium At Equilibrium C3 Water Before Reaction CAa Water HCI At Equilibrium CAb Water NaOH Before Reaction At Equilibrium

Explanation / Answer

1. Buffer solution: First calculate the molarities of NaC2H3O2 and HC2H3O2

NaC2H3O2.3H2O = 3.3g*1mol/136.08g/0.050L = 0.49M

HC2H3O2 = 6.0M*4mL/50mL = 0.48M

Now using Ka for acetic acid, we can calculate the [H+] concentration.

Ka = [H3O+][CH3COO-] / [CH3COOH]

So, [H3O+] = Ka [CH3COOH]/CH3COO-] = 1.8*10-5 * 0.48M/0.49M = 1.76*105

pH = - log[H+] = - log(1.76*10-5) = 4.75

2. Buffer + HCl

50mL buffer + 1ml 3.0M HCl

So, [HCl] = 3.0M*1mL/51mL = 0.059M

By setting the ICE table, we can calculate the equilibrium concentrations.

CH3COOH H3O+ CH3COO-

Initial 0.48M 0.059M 0.49M

change +0.059 ---- -0.059

equilibrium 0.539M x 0.431M

The added protons from HCl combine with the acetate ions to form more acetic acid.

Substitute these values from table in Ka expression,

[H3O+] = x = 1.8*10-5(0.539M/0.431M) = 2.25*10-5

pH = -log(2.25*10-5) = 4.65

3. Buffer + NaOH

50mL buffer + 1mL 3.0M NaOH

[NaOH] = 3.0M*1mL/51mL = 0.059M

Again set up an ICE table to get the equilibrium concentrations

CH3COOH H3O+ CH3COO-

initial 0.48M ----- 0.49M

change - 0.059 ------ +0.059

equilibrium 0.421M x 0.549M

So, [H+] = x= 1.8*10-5 (0.421M/0.549M) = 1.38*10-5

pH = -log(1.38*10-5) = 4.86

4. water

for a neutral solution, [H+] = 10-7

So, pH = 7

5. water + HCl

[HCl] = 0.059M

So, pH = -log(0.059) = 1.2

In the absence of HC2H3O2 and C2H3O2- , the same concentration of HCl will produce a pH of 1.2

6. water + NaOH

[NaOH] = 0.059M

pOH = - log 0.059 = 1.2

so, pH = 14 - 1.2 = 12.8 (without buffer)

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