I need help figuring this out, not just the answer please. I know he lays it out

Question

I need help figuring this out, not just the answer please. I know he lays it out for us pretty well, but I'm still getting stuck. Thank you in advance!

From the following information, determine the saponification number:

* Sample weight of oil = 0.75 grams

* mL of approximately 0.5 N KOH employed = 20 mL

* mL of standard 0.1 N HCl to titrate the KOH blank = 95 mL

* mL of 0.1 N HCl to titrate the saponified mixture = 25 mL

* MW (and EW) of KOH = 56.12 g/mol

* MW (and EW) of HCl = 36.46 g/mol

We were given hints to figure it out and I think I have done so correctly up until hint four - the part where we are to find the ratio equation for KOH. The hints are as follows:

1. start with finding g of KOH in 20mL of 0.5N KOH

2. next find the grams of HCl in 95mL of 0.1N HCl. This is the amount of grams that reacts with the amount of grams you found above in step 1 without any fats/oils involved.

3. again find the grams of HCl in 25mL of 0.1N HCl. This is the amount of grams that reacts with free KOH after the KOH has reacted with the oil.

4. set up a ratio equation to find the grams of free KOH (that reacted with the grams you found in step 3)

5. find the number of grams that reacted with the oil by taking your step 1 answer - step 4 answer

6. use step 5 answer (use in mg not g) with the g of oil you have to find the saponificiation number

Explanation / Answer

Ans ie very simple

First find tha amount of KOH and HCl in grams then apply this formula

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