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You prepared the buffer by adding about 0.2 g of each of the weak acid and its conjugate base to 25.0 mL of water. The goal was to produce a buffer near its point of maximum buffer capacity where [WA] = [CB]. An alternative approach could have been to dissolve about 0.4 g of the weak acid in water, and then add enough NaOH to convert half the weak acid into its conjugate base. How many mL of 0.10 M NaOH would have been needed to achieve this result?

Respuesta :

ajeigbeibraheem

Answer:

8.55 mL of NaOH is needed to achieve the result

Explanation:

Given that ;

the mass of the weak acid = 0.4 g

molecular weight = 234 g/mol

therefore number of moles of the weak acid = 0.4 g/234 g/mol = 0.00171 mole

We need to convert half the weak acid (WA) to conjugate base  (CB) by adding NaOH.

[WA]=[CB] 0.00171/2 = 8.55×10⁻⁴  mole needed

Also ; given that moles of NaOH = 0.10 M

Then the number of mL needed to achieve this result =

[tex]8.55*10^{-4}*\frac{1 \ L}{0.1 \ M}[/tex]

= [tex]8.55*10^{-4}*\frac{1 000}{0.1 }[/tex]

= 8.55 mL of NaOH is needed to achieve the result

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