Question

35. What is the pH for a 0.80M HCSH5NCI solution? Kb for CsHsN is 1.7 x 10 Fnd ka SN 36, What is the pH for a 0.050 M Sodium Benzoate (NaCHsO2) solution? Ka for HC7H502 is 6.4 x 10 weat /st

Answer 1

35)

Ionization equilibrium of HC5H5⁺ is

HC5H5N⁺(aq) + H2O(l) <--------> C5H5N(aq) + H3O⁺(aq)

Ka= [C5H5N][H3O⁺] /[HC5H5N⁺]

Ka= Kw/Kb = 1.00 ×10^-14/1.7 ×10^-9 = 5.88 ×10^-6

at equilibrium

[HC5H5N⁺] = 0.80 - x

[ C5H5N] = x

[H3O⁺] = x

so,

x²/ ( 0.80 - x) = 5.88 ×10^-6

solving for x

x = 0.002166

[H3O⁺] = 0.002166M

pH = -log[H3O⁺]

pH = - log(0.002166)

pH = 2.66

36)

Ionization equilibrium of C7H5O2^- is

C7H5O2^-(aq) + H2O(l) <------> HC7H5O2(aq) + OH^-(aq)

Kb = [HC7H5O2][OH^-]/[C7H5O2^-]

K_{​​​​​​}b = Kw/Ka = 1.00 ×10^-14/6.4×10^-5 = 1.56×10^-10

at equilibrium

[C7H5O2^-] = 0.050 - x

[HC7H5O2^-] = x

[OH^-] = x

so,

x²/(0.050 - x) = 1.56 ×10^-10

solving for x

x = 2.793×10^-6

[OH^-] = 2.793 ×10^-6M

pOH= -log[OH^-]

pOH = -log(2.793 ×10^-6)

pOH = 5.56

pH = 14 - 5.56

pH = 8.44