a) Explain how you would prepare a buffer with a pH = 7.2 using carbonic acid H2

Question

a) Explain how you would prepare a buffer with a pH = 7.2 using carbonic acid H2CO3 and NaHCO3 solutions. What are concentrations and volumes of the two solutions you would use? (Ka1 of H2CO3 is 4.5 x 10–7 )

b)Aqueous benzoic acid (C6H5COOH, Ka = 6.3 x10-5 ) solution dissociates into hydronium ions and benzoate anions. At 300 ºC, 15.00 mL of 0.0120 M of sodium benzoate (NaC6H5COO) are added to a 55.00 mL of 0. 0250 M benzoic acid and equilibrium is reestablished, what is the concentration of the hydronium ion? the pH of the resulting solution?

C6H5COOH(aq) <==> H3 +O(aq) + C6H5COO- (aq)

Explanation / Answer

(a): We can calculate the ratio of concentration of H2CO3 and NaHCO3 by applying Hendersen equation.

pH = pKa + log [NaHCO3] / [H2CO3]

=> 7.2 = - log(4.5x10-7) + log[NaHCO3] / [H2CO3] = 6.347 + log[NaHCO3] / [H2CO3]

=> log[NaHCO3] / [H2CO3] = 0.853

=> [NaHCO3] / [H2CO3] = 100.853 = 7.13

or (moles of NaHCO3) / (mole of H2CO3) = 7.13

Hence the moles of NaHCO3 used should be 7.13 times higher than the moles of H2CO3 to get a pH of 7.2

(b): moles of C6H5COOH in the solution = MxV = 0.0250 mol/L x 0.055 L = 0.001375 mol

moles of C6H5COONa in the solution = MxV = 0.0120 mol/L x 0.015 L = 0.00018 mol

Now we can calculate the pH by applying Hendersen equation.

pH = pKa + log[C6H5COONa] / [C6H5COOH]

=> pH = - log(6.3x10-5) + log(mles of C6H5COONa / moles of C6H5COOH) [ Since volume is same for both]

=> pH = 4.20 + log (0.00018 mol / 0.001375 mol)

=> pH = 3.32 (answer)

=> - log[H+] = 3.32

=> [H+] = 10-3.32 = 4.81x10-4 M (answer)

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