When 32.40 mL of 0.258 M NaOH is used to titrate a 1.00 mL sample, the concentration of acetic acid in the mixture is 8.36 M.
Let's consider the neutralization reaction between sodium hydroxide and acetic acid.
NaOH + CH₃COOH ⇒ CH₃COONa + H₂O
32.40 mL of 0.258 M NaOH react. The reacting moles of NaOH are:
[tex]0.03240 L \times \frac{0.258mol}{L} = 8.36 \times 10^{-3} mol[/tex]
The molar ratio of NaOH to CH₃COOH is 1:1. The moles of CH₃COOH required to react with 8.36 × 10⁻³ moles of NaOH are:
[tex]8.36 \times 10^{-3} mol NaOH \times \frac{1molCH_3COOH}{1molNaOH} = 8.36 \times 10^{-3} mol CH_3COOH[/tex]
8.36 × 10⁻³ moles of CH₃COOH are in 1.00 mL of solution. The concentration of CH₃COOH is:
[tex]M = \frac{8.36 \times 10^{-3} mol}{1.00 \times 10^{-3} L} = 8.36 M[/tex]
When 32.40 mL of 0.258 M NaOH is used to titrate a 1.00 mL sample, the concentration of acetic acid in the mixture is 8.36 M.
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Titration is the laboratory process in which concentration of a substance is determined by the addition of titrant. The concentration of acetic acid in the mixture is 8.36 M.
Given that,
32.40 mL of 0.258 M NaOH is used to titrate a 1.00 mL sample.
The reaction between sodium hydroxide and acetic acid can be considered as neutralization reaction.
The given reaction can be represented as:
Now, 32.40 mL of 0.258 M NaOH reacts, such that reacting moles of NaOH are:
Thus, [tex]8.36\times 10^{-3\;}\text{mol}[/tex]es of CH[tex]_3[/tex]COOH are present in 1.00mL of solution, such that the concentration of CH[tex]_3[/tex]COOH will be:
[tex]\text M&=\dfrac{8.36\times10^{-3}}{1.00\times10^{-3}}&=8.36\;\text M[/tex]
Therefore, 32.40 mL of 0.258 M NaOH is used for the titration of 1.00mL sample, in which the concentration of acetic acid in the mixture is 8.36 M.
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