A. A solution of equal molar concentrations of the artificial sweetener sacchari

Question

A. A solution of equal molar concentrations of the artificial sweetener saccharin and its sodium salt was found to have pH = 3.08. (i) What are the values of pKa and Ka of saccharin? (ii) What would the pH be if the concentration of acid was twice that of the salt?

B. A buffer solution of volume 100.0 mL is 0.140 m Na2HPO4(aq) and 0.120 m KH2PO4(aq). (iii) What are the pH and the pH change resulting from the addition of 75.0 mL of 0.0100 m NaOH(aq) to the buffer solution? (iiii) What are the pH and the pH change resulting from the addition of 10.0 mL of 0.50 m HNO3(aq) to the initial buffer solution?

Explanation / Answer

A.

pH = pka + log [salt/acid]

if salt and acid are present in equal concentration then pH = pka.

pka of saccarin = 3.08

Ka = 10^-pka = 10^-3.08 = 8.32*10^-4

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