44. The pH of a solution which is 0.1 M in HA and
0.5 M in NaA. Kₐ for HA is 1.8x10⁻⁶
(a) 5.44 (b) 6.44
(c) 6.0 (d) 4.73
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(b) 6.44
Explanation:
Given: 0.1 M in HA and 0.5 M in NaA. Kₐ for HA is 1.8x10⁻⁶
pH of a solution = PKa + log salt/acid
= -log(1.8 * 10^-6) + 0.5/0.1
= 5.74 + 0.6989
= 6.44
Option B is the answer.
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