Answer to Question #177177 in General Chemistry for Annie

Question

1.047g of NAHCO3 and 2.473g of NA2CO3 are dissolved in the same beaker of water and transferred to a volumetric flask and made to 100ml. The Ka for HCO3- is 4.7 x 10^ -11

a) Calculate the theoretical pH pf the final solution if 10ml of 0.05 mol L HCL was added to 50ml of this buffer solution.

b) Calculate the pH of the resulting solution if 0.02g of solid sodium hydroxide was added to 50ml of this buffer solution

Answer 1

Q177177

1.047g of NaHCO₃ and 2.473g of Na₂CO₃ are dissolved in the same beaker of water and transferred to a volumetric flask and made to 100ml. The Ka for HCO₃- is 4.7 x 10^ -11

a) Calculate the theoretical pH of the final solution if 10ml of 0.05 mol L HCL was added to 50ml of this buffer solution.

b) Calculate the pH of the resulting solution if 0.02g of solid sodium hydroxide was added to 50ml of this buffer solution.

Solution:

Step 1 : To find the concentration of NaHCO3 and Na2CO3 in the buffer solution.

Let us first find the concentration of NaHCO₃ and Na₂CO₃.

1.047 g of NaHCO₃ and 2.473 g of Na₂CO₃ were mixed in water and diluted to 100 mL in a volumetric flask.

Molar mass of NaHCO₃ = 84.006 g/mol

Molar mass of Na₂CO₃ = 105.99 g/mol

100mL in 'L' = 100 mL * 1 L / 1000mL = 0.100 L ;

$$moles\space of\space NaHCO_3 = 1.047 g\space of\space NaHCO_3 * \frac{1mol\space NaHCO_3}{84.006 \space g\space of\space NaHCO_3}$$

$= 0.01246\space mol\space of\space NaHCO_3$

$$moles\space of\space Na_2CO_3 = 2.473 g\space of\space Na_2CO_3 * \frac{1mol\space Na_2CO_3}{105.99 \space g\space of\space Na_2CO_3}$$

$$= 0.02333\space mol\space of\space Na_2CO_3$$

Next, find the molarity of NaHCO₃ and Na₂CO₃

$$molarity\space of\space NaHCO_3 = \frac{mol \space of\space NaHCO_3 }{volume\space of\space solution\space in\space 'L' }$$

$$molarity\space of\space NaHCO_3 = \frac{0.01246\space mol \space of\space NaHCO_3 }{0.100L }$$

$$molarity\space of\space NaHCO_3 = 0.1246\space mol/L$$

$$molarity\space of\space Na_2CO_3 = \frac{mol \space of\space Na_2CO_3 }{volume\space of\space solution\space in\space 'L' }$$

$$molarity\space of\space Na_2CO_3 = \frac{0.02333\space mol \space of\space Na_2CO_3 }{0.100L }$$

$$molarity\space of\space Na_2CO_3 = 0.2333\space mol/L$$

Step 2 : 10 mL of HCl is added to 50 mL of the buffer solution.

Let us use the dilution formula and find the new concentration of HCl, NaHCO3, and Na2CO3 just after mixing.

For NaHCO₃

$M_fV_f = M_i V_i$

$$M_f * 60 mL = 0.1246 M * 50 mL$$

$$M_f (NaHCO_3 ) = 0.1038 M.$$

For Na₂CO₃

$M_fV_f = M_i V_i$

$$M_f * 60 mL = 0.2333 M * 50 mL$$

$$M_f (Na_2CO_3 ) = 0.1944 M.$$

For HCl

$$M_fV_f = M_i V_i$$

$$M_f * 60 mL = 0.05 M * 10 mL$$

$$M_f (HCl ) = 0.00833 M.$$

Now we will prepare the ICE Table and find the equilibrium concentration of NaHCO₃ and Na₂CO₃

In our given buffer Na₂CO₃ is the base component and NaHCO₃ is the acid component.

HCl reacts with the base component (Na₂CO₃ ) to form NaHCO₃. Cl- is the spectator ion, so

I have not shown it in the given reaction.

Ka of HCO₃- = 4.7 * 10^-11 .

pKa = -log [ 4.7 * 10^-11 ] = 10.328

Next we will use the Henderson - Hasselbalch equation and find the pH of the solution.

plug pKa = 10328, [ Na₂CO₃ ]= 0.1861M , [NaHCO₃ ] = 0.1121 M

$$pH = pKa + log [\frac{[Base]}{[Acid]}] = pKa + log [\frac{[A^-]}{[HA]}]$$

$$pH = pKa + log[\frac{[[Na_2CO_3]}{[NaHCO_3]}]$$

$$pH = 10.328 + log[\frac{[0.1861M}{0.1121M}]$$

$$pH = 10.328 + log[1.6601] = 10.328 + 0.2201 = 10.548 = 10.55$$

Answer a) The pH of the buffer solution after the addition of 10 mL of 0.05 M HCl will be 10.55.

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Part B: Calculate the pH of the resulting solution if 0.02g of solid sodium hydroxide was added to 50ml of this buffer solution.

0.02 g of solid NaOH is added to 50 mL of the buffer solution.

We will assume that there is no change in volume on the addition of 0.02 g of NaOH to the 50 mL buffer.

We will use concentrations if NaHCO₃ and Na₂CO₃ which we calculated in Step 1.

[NaHCO₃ ] = 0.1246 mol/L

[Na₂CO₃] = 0.2333 mol / L

Molar mass of NaOH = 39.9969 g/mol.

50 mL in 'L' = 50 mL * 1L/1000 mL = 0.050 L

$$moles\space of\space NaOH = 0.02\space g \space NaOH * \frac{1\space mol\space NaOH }{39.9969\space g\space of\space NaOH }$$

$moles\space of\space NaOH = 0.000500\space mol \space NaOH$

$$concentration\space of\space NaOH = \frac{0.000500\space mol\space NaOH }{0.050\space L }$$

$$concentration\space of\space NaOH = 0.01000 mol/L$$

So now we have

[NaHCO₃ ] = 0.1246 mol/L

[Na₂CO₃] = 0.2333 mol / L

[NaOH] = 0.01000 mol/L .

These are the initial concentration of the species just after the addition of NaOH.

We will draw the ICE Table and find the equilibrium concentration of all the species.

Na₂CO₃ is the base and NaHCO₃ is acid, NaOH will react with the acid component of the buffer.

For the ICE table, at equilibrium we have

[NaHCO₃ ] = 0.1146 mol/L

[Na₂CO₃] = 0.2433 mol / L

plug this in the Henderson -Hasselbalch equation and find the pH of the buffer.

plug pKa = 10328, [ Na₂CO₃ ]= 0.2433M , [NaHCO₃ ] = 0.1146 M

$$pH = pKa + log [\frac{[Base]}{[Acid]}] = pKa + log [\frac{[A^-]}{[HA]}]$$

$$pH = pKa + log[\frac{[[Na_2CO_3]}{[NaHCO_3]}]$$

$$pH = 10.328 + log[\frac{[0.2433M}{0.1146M}]$$

$$pH = 10.328 + log[2.123] = 10.328 + 0.3270 = 10.655 = 10.65$$

Hence the pH of the buffer after the addition of 0.02 grams of NaOH is 10.65.