Hello all! I need to make 100 mL of the following solution. I tried to use the C1V1 formula, but I have dyscalculia so my math skills are non-existent.
Ponceau-stained membrane can be detained using:
Distilled water.
5% acetic acid solution.
0.1M NaOH solution.
0.1M solution of NaOH would contain 0.1 mole of NaOH per liter of water. The molar mass of NaOH is 40 grams/mole. O.1 mole would therefore be 4 grams of NaOH. Dissolve 4 grams of NaOH in 1 liter of water to obtain 1 liter of a 0.1M NaOH solution.
Then in 100 ML =0.4 grams in 100 ML.
5 % acetic acid means 95 ML of distilled water.
Simply you can take 0.4 gram of NaOH and mixed 95 ML distilled water add 5 ML acetic acid.
Now, measure 5 mL of acetic acid, pour it into the 100 mL standard flask and mixed 0.4 gram of NaOH, and top it up to the mark with distilled water.
Regards