MCP๋กœ ์—ฐ๊ฒฐ โ†’

๊ณ„์‚ฐ ์ž…๋ ฅ

๊ณต์‹

๊ด‘๊ณ 

๊ฒฐ๊ณผ

๋น„๋ˆ„ํ™” ๊ฐ’
200.36
์ง€๋ฐฉ 1g๋‹น mg KOH
KOH ๋ชฐ์ˆ˜ 0.05 mol
์ง€๋ฐฉ ์งˆ๋Ÿ‰ 14 g
์ ์šฉ๋œ KOH ๋ชฐ์งˆ๋Ÿ‰ 56,100 mg/mol

๋น„๋ˆ„ํ™” ๊ฐ’์ด๋ž€?

๋น„๋ˆ„ํ™” ๊ฐ’(SAP ๊ฐ’)์€ ์ง€๋ฐฉ์ด๋‚˜ ๊ธฐ๋ฆ„ 1g์„ ๋น„๋ˆ„ํ™”ํ•˜๋Š” ๋ฐ ํ•„์š”ํ•œ ์ˆ˜์‚ฐํ™”์นผ๋ฅจ(KOH)์˜ ์–‘์„ ๋ฐ€๋ฆฌ๊ทธ๋žจ์œผ๋กœ ๋‚˜ํƒ€๋‚ธ ์ˆ˜์น˜์ž…๋‹ˆ๋‹ค. ์ด ๊ฐ’์€ ์‹œ๋ฃŒ์— ๋“ค์–ด ์žˆ๋Š” ์ง€๋ฐฉ์‚ฐ์˜ ํ‰๊ท  ๋ถ„์ž๋Ÿ‰, ์ฆ‰ ํ‰๊ท  ์‚ฌ์Šฌ ๊ธธ์ด๋ฅผ ๊ฐ€๋Š ํ•˜๋Š” ์ค‘์š”ํ•œ ์ง€ํ‘œ์ž…๋‹ˆ๋‹ค. ๋น„๋ˆ„ํ™” ๊ฐ’์ด ๋†’์„์ˆ˜๋ก ์‚ฌ์Šฌ์ด ์งง์€ ์ง€๋ฐฉ์‚ฐ์ด ๋งŽ๊ณ , ๊ฐ’์ด ๋‚ฎ์„์ˆ˜๋ก ์‚ฌ์Šฌ์ด ๊ธด ์ง€๋ฐฉ์‚ฐ์ด ๋งŽ๋‹ค๋Š” ๋œป์ž…๋‹ˆ๋‹ค. ํ™”ํ•™์ž, ๋น„๋ˆ„ ์ œ์กฐ์ž, ์‹ํ’ˆ ๊ณผํ•™์ž๊ฐ€ ์ง€๋ฐฉ๊ณผ ๊ธฐ๋ฆ„์˜ ํŠน์„ฑ์„ ๋ถ„์„ํ•  ๋•Œ ๋‘๋ฃจ ํ™œ์šฉํ•ฉ๋‹ˆ๋‹ค.

Diagram of a triglyceride molecule reacting with three KOH to form glycerol and three soap molecules
Saponification: a fat (triglyceride) reacts with KOH to yield glycerol and soap (fatty acid salts).

๊ณ„์‚ฐ๊ธฐ ์‚ฌ์šฉ ๋ฐฉ๋ฒ•

์‹œ๋ฃŒ์™€ ๋ฐ˜์‘ํ•˜์—ฌ ๋น„๋ˆ„ํ™”์— ์“ฐ์ธ KOH์˜ ๋ชฐ์ˆ˜์™€, ์ง€๋ฐฉ ๋˜๋Š” ๊ธฐ๋ฆ„์˜ ์งˆ๋Ÿ‰(g)์„ ์ž…๋ ฅํ•˜์„ธ์š”. ๊ณ„์‚ฐ๊ธฐ๋Š” KOH ๋ชฐ์ˆ˜์— 56,100 mg/mol(KOH์˜ ๋ชฐ์งˆ๋Ÿ‰์„ ๋ฐ€๋ฆฌ๊ทธ๋žจ์œผ๋กœ ํ™˜์‚ฐํ•œ ๊ฐ’)์„ ๊ณฑํ•œ ๋’ค ์‹œ๋ฃŒ ์งˆ๋Ÿ‰์œผ๋กœ ๋‚˜๋ˆ„์–ด, ์ง€๋ฐฉ 1g๋‹น mg KOH ๋‹จ์œ„์˜ SAP ๊ฐ’์„ ๊ตฌํ•ฉ๋‹ˆ๋‹ค.

๊ณต์‹ ํ’€์ด

๋น„๋ˆ„ํ™” ๊ฐ’(mg KOH/g) = (56100 ร— KOH ๋ชฐ์ˆ˜) รท ์ง€๋ฐฉ ์งˆ๋Ÿ‰(g). ์ƒ์ˆ˜ 56,100์€ KOH์˜ ๋ชฐ์ˆ˜๋ฅผ ๋ฐ€๋ฆฌ๊ทธ๋žจ์œผ๋กœ ๋ฐ”๊ฟ” ์ค๋‹ˆ๋‹ค(KOH์˜ ๋ชฐ์งˆ๋Ÿ‰์€ ์•ฝ 56.1 g/mol = 56,100 mg/mol). ์‹œ๋ฃŒ ์งˆ๋Ÿ‰์œผ๋กœ ๋‚˜๋ˆ„๋ฉด ๊ฒฐ๊ณผ๊ฐ€ ์ง€๋ฐฉ 1g ๊ธฐ์ค€์œผ๋กœ ์ •๊ทœํ™”๋ฉ๋‹ˆ๋‹ค.

$$\text{SAP} = \frac{56100 \times \text{Moles of KOH (mol)}}{\text{Mass of Fat / Oil (g)}}$$
Formula breakdown showing SAP equals 56100 times moles KOH divided by fat mass
The SAP value scales the moles of KOH per gram of fat by the molar mass of KOH (56100 mg/mol).

๊ณ„์‚ฐ ์˜ˆ์‹œ

์˜ˆ๋ฅผ ๋“ค์–ด ์–ด๋–ค ๊ธฐ๋ฆ„ 14g์„ ๋น„๋ˆ„ํ™”ํ•˜๋Š” ๋ฐ KOH 0.05 mol์ด ํ•„์š”ํ•˜๋‹ค๊ณ  ๊ฐ€์ •ํ•ด ๋ด…์‹œ๋‹ค. SAP = (56100 ร— 0.05) รท 14 = 2805 รท 14 โ‰ˆ 200.36 mg KOH/g์ž…๋‹ˆ๋‹ค.

$$\text{SAP} = \frac{56100 \times 0.05}{14} = \frac{2805}{14} \approx 200.36\ \text{mg KOH/g}$$

์ด๋Š” ์ค‘๊ฐ„ ๊ธธ์ด ์ง€๋ฐฉ์‚ฐ์œผ๋กœ ์ด๋ฃจ์–ด์ง„ ํŠธ๋ฆฌ๊ธ€๋ฆฌ์„ธ๋ฆฌ๋“œ์—์„œ ํ”ํžˆ ๋‚˜ํƒ€๋‚˜๋Š” ๊ฐ’์ž…๋‹ˆ๋‹ค.

์ž์ฃผ ๋ฌป๋Š” ์งˆ๋ฌธ

KOH์˜ ๋ชฐ์งˆ๋Ÿ‰์€ ์™œ 56,100์ธ๊ฐ€์š”? KOH์˜ ๋ชฐ์งˆ๋Ÿ‰์€ ์•ฝ 56.1 g/mol์ž…๋‹ˆ๋‹ค. ์ด๋ฅผ 1๋ชฐ๋‹น ๋ฐ€๋ฆฌ๊ทธ๋žจ์œผ๋กœ ํ™˜์‚ฐํ•˜๋ฉด 56,100 mg/mol์ด ๋˜๋ฉฐ, ๋•๋ถ„์— SAP ๊ฐ’์ด ํ‘œ์ค€ ๋‹จ์œ„์ธ mg KOH/g์œผ๋กœ ์œ ์ง€๋ฉ๋‹ˆ๋‹ค.

KOH ๋Œ€์‹  NaOH๋ฅผ ์จ๋„ ๋˜๋‚˜์š”? ์ „ํ†ต์ ์ธ ๋น„๋ˆ„ํ™” ๊ฐ’์€ KOH๋ฅผ ๊ธฐ์ค€์œผ๋กœ ์ •์˜๋ฉ๋‹ˆ๋‹ค. ๋งŒ์•ฝ NaOH(40 g/mol)๋กœ ์ธก์ •ํ–ˆ๋‹ค๋ฉด, ๊ฒฐ๊ณผ์— \(56.1/40 \approx 1.4025\)๋ฅผ ๊ณฑํ•ด KOH ๊ธฐ์ค€ ๊ฐ’์œผ๋กœ ํ™˜์‚ฐํ•˜๋ฉด ๋ฉ๋‹ˆ๋‹ค.

์ผ๋ฐ˜์ ์ธ ๋น„๋ˆ„ํ™” ๊ฐ’์€ ์–ด๋А ์ •๋„์ธ๊ฐ€์š”? ๋Œ€๋ถ€๋ถ„์˜ ์‹๋ฌผ์„ฑ ๊ธฐ๋ฆ„์€ ๋Œ€๋žต 180~200 mg KOH/g ์‚ฌ์ด์— ์†ํ•ฉ๋‹ˆ๋‹ค. ์ฝ”์ฝ”๋„› ์˜ค์ผ์€ ์‚ฌ์Šฌ์ด ์งง์€ ์ง€๋ฐฉ์‚ฐ์ด ๋งŽ์•„ ์•ฝ 250์œผ๋กœ ๋” ๋†’๊ฒŒ ๋‚˜ํƒ€๋‚ฉ๋‹ˆ๋‹ค.

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