1) phenol (C6H5OH) is a weak acid. The pH of a 0.010M solution of sodium phenola

Question

1) phenol (C6H5OH) is a weak acid. The pH of a 0.010M solution of sodium phenolate (NaOC6H5) is 11.00. What is the acid ionization constant of phenol?
2) A solution of 0.23 mol of the hydrochloride salt of quintine, a weak organic base used in the treatment of malaria, in enough water to form 1.00L of solution has a pH of 4.58. Calculate Kb of quintine. 1) phenol (C6H5OH) is a weak acid. The pH of a 0.010M solution of sodium phenolate (NaOC6H5) is 11.00. What is the acid ionization constant of phenol?
2) A solution of 0.23 mol of the hydrochloride salt of quintine, a weak organic base used in the treatment of malaria, in enough water to form 1.00L of solution has a pH of 4.58. Calculate Kb of quintine.
2) A solution of 0.23 mol of the hydrochloride salt of quintine, a weak organic base used in the treatment of malaria, in enough water to form 1.00L of solution has a pH of 4.58. Calculate Kb of quintine.

Explanation / Answer

C6H5O- + H2O (l) <---> C6H5OH (aq) + OH-(aq)

given pH = 11 , pOH = 14-11 = 3 , [OH-] = 10^-pOH = 10^-3 M

at equilibrium [C6H5OH] = [OH-] = 10^-3 , [C6H5O-] = ( 0.01-10^-3) = 0.009

Kb = [OH-][C6H5OH] /[C5H5O0] = ( 10^-3) ( 10^-3) / ( 0.009) = 0.00011

Ka = acid ionization constant = Kw/Kb = ( 10^-14) / ( 0.00011) = 9 x10^ -11

2) HCl salt of qunitine is acidic if quintie is base , let the salt be represented as QH+ Cl- ( Cl- is ignored as it is spectator ion)

QH+ (aq) <----> Q + H+ (aq)

pH = 4.58 , [H+] = 10^ -4.58 = 2.63 x 10^ -5 M = [Q]

Ka = [H+][Q]/[QH+] = ( 2.63x10^-5) ( 2.63x10^ -5) / ( 0.23) = 3 x10^ -9

Kb of Q= Kw / Ka ( QH+) = 10^ -14 / ( 3x10^-9) = 3.33 x 10^ -6

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