How many mL of a 0.100 M NaOH solution would be needed to titrate 0.156 g of butanoic acid to a neutral pH endpoint?
Answer:
Butanoic acid weight = 0.156g
Molecular weight of Butanoic acid = 88
Number of moles of Butanoic acid = weight / molar mass
= 0.156 / 88
=0.00177
We know that Molarity = Number of moles / Volume in liter
Assume Butanoic acid volume (V1) = 1 liter
Butanoic acid concentration for one liter (M1) = 0.00177 M
NaOH concentration (M2) = 0.1 M
NaoH volume (V2) = ?
We know that M1V1 = M2V2
0.00177 x 1 = 0.1 x V2
V2 = 0.00177/ 0.1
V2 = 0.0177 liter
Hence for 1 liter of Butanoic acid 0.0177 liter or 17.7 mL of NaOH is required.
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