hi
succinic acid is a diprotic acid, it releases one proton from the HOOC- and one proton from the -COOH
for neutralisation we will need: HOOCCH2CH2COOH + 2KOH
for every 1 mol of succinic acid we will need 2 mol of KOH
so 5 * 10⁻³ mol succinic acid need 1 * 10⁻² mol KOH
V=n/C = 1 * 10⁻² / 15 * 10⁻² = 1/15 = 0,0666 L = 66,7 mL
is it possible for you to explain how you did that calculation? i understood the moles ratio but if possible, could you solve it on paper or write in fraction form
thanks
i use ari’s stochimetry table to solve
i knew the ratio of succinic acid to KOH was 1 : 2 so the molar ratio had to be 5 * 10⁻³ : 2 * 5 * 10⁻³
here’s a link to the video
https://imat.entermedschool.com/t/marathon-mrt-39-bmat-2017-q26-important-using-the-stochio-table-one-last-time-9-01/2872?u=juliette
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