Which of the following acids is the stronger, boric acid, pKa = 9.0 or acetic ac

Question

Which of the following acids is the stronger, boric acid, pKa = 9.0 or acetic acid, pKa = 4.6? Explain your answer with proper justification. Calculate the pH of a 0.2 M acetate buffer (a solution containing 0.1 mol L^-1 acetic acid and 0.1 mol L^-1 sodium acetate), given that the pKa of acetic acid is 4.76. What would be the pH value after adding 0.05 mmol of NaOH to 1 L of 0.2 mol L^-1 acetate buffet? Compare the latter pH value (in part 'b') with that obtained after adding 0.05 mmol NaOH to 1 L of water (a simple solution of 0.05 mol L^-1 NaOH) in part 'c'. How would you explain the change in pH two scenarios?

Explanation / Answer

a) Acetic acid is stronger

pKa = -logKa

Ka = [Conjugate base][H+]/[Acid]

From the above relations , we understand that If Hydrogen ion concentration [H+] is high, then Ka value is high and pKa value will be less. So, less pKa value indicate stronger acid.

b) To solve this problem we have to use Henderson-Hasselbalch equation

pH = pKa + log([A-]/HA])

where, [A-] is concentration of conjucate base, 0.1M

[HA] is concentration of conjucate acid,0.1M

So , pH = 4.76 + log(0.1M/0.1M)

Therefore, pH = 4.76

c)before the addition of NaOH the equilibrium is

CH3COOH -----------> CHCOO- + H+

< ----------

0.00005M of NaOH it with 0.00005M of acetic acid

OH- + CH3COOH ----------> CH3COO- + H2OO

So, 0.00005mol of CH3COOH is Decreased and 0.00005mols of CH3COO- increased

Therefore, the new concentration of CH3COOH = 0.09995M

the new concentration of CH3Hessel 0.10005M

pKa = 4.76

Substituting the values in Henderson - Hesselbalch equation

pH = 4.76 +log(0.10005M/0.09995)

=4.7595 = 4.76

So, pH is not changing after addition of 0.00005 mol of NaOH

d) If 0.05mmol of NaOH added to 1litre of water

[OH-] = 0.00005mol in 1 litre = 0.00005M

pOH = -log[OH-] = - log[0.00005] = 4.30

pH + pOH = 14

Therefore, pH = 14 -4.30 = 9.70

If 0.00005 mol of NaOH added to 1 Litre of buffer solution ,pH of the buffer solution not varying

If 0.00005 mol of NaOH added to 1 litre of water then the pH of water raised to 7 to 9.7

This indicating the usefulness of buffer solutions.

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