What combination of substances will give a buffered solution that has a ph of 5.05? (assume each pair of substances is dissolved in 5.0 l of water.) (kb for nh3 = 1.8 *10^–5; kb for c5h5n = 1.7 * 10^–9)
a) 1.0 mole nh3 and 1.5 mole nh4cl
b) 1.5 mole nh3 and 1.0 mole nh4cl
c) 1.0 mole c5h5n and 1.5 mole c5h5nhcl
d) 1.5 mole c5h5n and 1.0 mole c5h5nhcl
e) none of these

C

Explanation:

When the Kb is given, the Henderson-Hasselbalch equation can be used to calculate the pOH of a buffer solution:

pOH = pKb + log ([A⁻] / [HA]) = -log(Kb) + log ([BH+] / [B])

Here, moles can be used in place of the concentration since the pairs listed are both dissolved in 5L, which cancel due to the fraction in the logarithm.

a) pOH = -log(1.8 x 10⁻⁵) + log(1.5/1.0) = 4.92

pH = 14 - pOH = 14 - 4.92 = 9.08

b) pOH = -log(1.8 x 10⁻⁵) + log(1.0/1.5) = 4.57

pH = 14 - pOH = 14 - 4.57 = 9.43

c) pOH = -log(1.7 x 10⁻⁹) + log(1.5/1.0) =  8.95

pH = 14 - pOH = 14 - 8.95 = 5.05

d) pOH = -log(1.7 x 10⁻⁹) + log(1.0/1.5) =  8.59

pH = 14 - pOH = 14 -  = 5.41

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