To show that you are justified in using [OH-] as a constant: using concentrations from Run 1, show that the concentration of NaOH has changed insignificantly when half of a CV chloride has been consumed?

Instructions
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Data table
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1 Answer
Jun 28, 2018

Answer:

The amount of NaOH reacted after one half-life is 0.02 % of the original.

Explanation:

Step 1. Calculate the initial concentration of CV

Your reaction beaker contained 15.00 mL of water, 5.00 mL of 0.05 mol/L NaOH, and 10.00 mL of the CV solution.

The reported values for the molar absorptivity of CV vary widely. The most precise value I found was #epsilon = "88 272.9 L·mol"^"-1""cm"^"-1"#.

We can use Beer's Law to calculate the concentration of CV in the reaction mixture.

#color(blue)(bar(ul(|color(white)(a)A = epsilonclcolor(white)(a)|)))" "#

where

#epsilon =# the molar absorptivity
#c =# the molar concentration
#l =# the path length in centimetres

I assume that the path length of your cuvette was 1 cm. Then

#c = A/(epsilon l) = 0.995/("88 279.2" "L·mol"^"-1""cm"^"-1" × 1 "cm") = 1.13 × 10^"-5" color(white)(l)"mol/L"#.

Step 2. Calculate the initial moles of CV

#"Moles of CV "= "0.010 00" color(red)(cancel(color(black)("L CV"))) × (1.13 × 10^"-5" color(white)(l)"mol CV")/(1 color(red)(cancel(color(black)("L CV"))))#

#= 1.13 × 10^"-7" color(white)(l)"mol CV"#

Step 3. Calculate the initial moles of #"NaOH"#

#"Moles of NaOH "= 0.005 color(red)(cancel(color(black)("L NaOH"))) × "0.05 mol NaOH"/( 1 color(red)(cancel(color(black)("L NaOH"))))#

#= 2.5 ×10^"-4"color(white)(l)"mol NaOH"#

After one half-life

The equation for the reaction is

#"CV"^"+" + "OH"^"-" → "CVOH"#

#"Moles of CV reacted" = (1.13 × 10^"-7" color(white)(l)"mol") /2 = 5.63 × 10^"-8"color(white)(l)"mol"#

#"Moles of NaOH reacted" = 5.63 × 10^"-8" color(red)(cancel(color(black)("mol CV"))) × "1 mol NaOH"/(1 color(red)(cancel(color(black)("mol CV"))))#

#= 5.63 × 10^"-8"color(white)(l)"mol NaOH"#

#"% NaOH reacted" = (5.63 × 10^"-8" color(red)(cancel(color(black)("mol NaOH"))))/(2.5 ×10^"-4" color(red)(cancel(color(black)("mol NaOH")))) × 100 % = 0.02 %#