Solution :
Part A) The pOH is calculated from Hendersen -Hasselbalch equation as,
pOH = pkb + log [Salt ] / [Base]
pkb is calculated as,
pkb = - log kb = - log 4.37 x 10^-4 = 4 - log 4.37
pkb = 4 - 0.64 = 3.36
Thus,
pOH = 3.36 + log (0.20 / 0.15)
pOH = 3.36 + 0.12 = 3.48
Hence,
pH = 14 - pOH = 14 - 3.48 = 10.52
Part B)
Addition of NaOH increases the concentration of base while decreases the concentration of salt.
Suppose concentration of NaOH = X mol/L
Thus,
For pH = 11, the pOH = 3
Therefore,
pOH = pkb + log (0.20 -X) / (0.15 +X)
3 = 3.48 + log (0.20 -X) / (0.15 + X)
log (0.20 -X) / (0.15+X) = - 0.48
log 0.20 - log X - log 0.15 - log X = - 0.48
log (0.20 / 0.15) - 2 log X = - 0.48
0.12 - 2 log X = - 0.48
2 log X = - 0.60
log X = - 0.30
X = antilog - 0.30 = 0.50 M
Hence, molarity of NaOH = 0.50 M
Since, total volume of buffer = 150 + 150 = 300 mL
Thus, using molarity equation.
Molarity = Number of moles / Volume in L
Number of moles = Molarity x Volume in L
= 0.50 M x 0.300 L = 0.15 mol
Hence,
Mass of NaOH = 0.15 mol x molar mass
= 0.15 M x 40 g mol-1 = 6.0 g of NaOH
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