Solution:
Number of moles of Na2CO3 = mass / molar mass
= 4.500 g / 106 g mol-1 = 0.0425 mol
Hence,
[Na2CO3] = number of mol / volume
= 0.0425 mol / 0.150 L
= 0.283 M
pH = pka + log [Na2CO3] / [NaHCO3]
9.50 = 10.25 + log 0.283 / [NaHCO3]
log [0.283] - log [NaHCO3] = - 0.75
log [NaHCO3] = 0.75 + log 0.283 = 0.75 - 0.55 = 0.20
[NaHCO3] = antilog 0.20 = 1.58 M
Hence, mass of NaHCO3
= Molarity x molar mass x volume in L
= 1.58 M x 84.0 g mol-1 x 0.150 L = 19.908 g
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