# 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 %
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 16, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://socratic.org/questions/to-show-that-you-are-justified-in-using-oh-as-a-constant-using-concentrations-fr", "fetch_time": 1695932526000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2023-40/segments/1695233510454.60/warc/CC-MAIN-20230928194838-20230928224838-00267.warc.gz", "warc_record_offset": 577979617, "warc_record_length": 7057, "token_count": 747, "char_count": 2175, "score": 4.34375, "int_score": 4, "crawl": "CC-MAIN-2023-40", "snapshot_type": "latest", "language": "en", "language_score": 0.7542018294334412, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00055-of-00064.parquet"}