Answer:
91.4%
Explanation:
Potassium hydrogen tartrate, KHT, reacts with KOH as follows.
KHT + KOH → H₂O + K₂T
where 1 mole of acid (KHT) reacts per mole o base (KOH), -That is acid/base ratio 1:1
The endpoint of a titration is the point in which moles of KOH = moles of KHT, you can see this endpoint with an indicator or doing a potentiometric titration.
As the endpoint requires 21.58mL = 0.02158L of a 0.1125M KOH, moles of KOH = moles of KHT are:
0.02158L × (0.1125mol / L) = 2.428x10⁻³ moles of KOH = moles of KHT
To convert these moles to grams you use molar mass of KHT (188.177g/mol):
2.428x10⁻³ moles of KHT × (188.177g / mol) = 0.457g of KHT are in the sample.
As you add 0.500g of sample, percentage of KHT in the sample is:
(0.457g / 0.500g) × 100 =
-That is the purity of the sample-