You have a vinegar solution you believe to be 0.83 M. You are going to titrate 20.0mL of it with a NaOH solution that you know to be 0.519 M. At what volume of added NaOH solution would you expect to see the end point?

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You have a vinegar solution you believe to be 0.83 M. You are going to titrate 20.0mL of it with a NaOH solution that you know to be 0.519 M. At what volume of added NaOH solution would you expect to see the end point?

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Answer 1 · 2017-05-08T23:47:14

You should expect to see the endpoint at 32 mL of added $NaOH$ $"NaOH"$ .

Explanation:

Let's write the formula for acetic acid as $HA$ $"HA"$ .

Then the equation for the reaction is

$HA + NaOH \to NaA + H_{2} O$ $"HA + NaOH" → "NaA" + "H"_2"O"$ .

1. Calculate the moles of $HA$ $"HA"$

$"Moles of HA" = 0.0200 color(red)(cancel(color(black)("L HA"))) ×( "0.83 mol HA")/(1 color(red)(cancel(color(black)("L HA")))) = "0.0166 mol HA"$

2. Calculate the moles of $"NaOH"$

$"Moles of NaOH" = 0.0166 color(red)(cancel(color(black)("mol HA"))) × ("1 mol NaOH")/(1 color(red)(cancel(color(black)("mol HA")))) = "0.0166 mol NaOH"$

3. Calculate the volume of the $"NaOH"$

$"Volume of NaOH" = 0.0166 color(red)(cancel(color(black)("mol NaOH"))) × ("1 L NaOH")/(0.519 color(red)(cancel(color(black)("mol NaOH")))) = "0.032 L" = "32 mL"$

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You have a vinegar solution you believe to be 0.83 M. You are going to titrate 20.0mL of it with a NaOH solution that you know to be 0.519 M. At what volume of added NaOH solution would you expect to see the end point?

1 Answer

Explanation:

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