# 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 Data table Jun 28, 2018

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 $\epsilon = \text{88 272.9 L·mol"^"-1""cm"^"-1}$.

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

$\textcolor{b l u e}{\overline{\underline{| \textcolor{w h i t e}{a} A = \epsilon c l \textcolor{w h i t e}{a} |}}} \text{ }$

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 $\text{NaOH}$

"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

$\text{CV"^"+" + "OH"^"-" → "CVOH}$

$\text{Moles of CV reacted" = (1.13 × 10^"-7" color(white)(l)"mol") /2 = 5.63 × 10^"-8"color(white)(l)"mol}$

"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 %