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How many mL of a 0.100 M NaOH solution would be needed to titrate 0.156 g...

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?

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Answer #1

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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