.500g of potassium hydrogen phthalate are titrated with .10 M sodium hydroxide.

Question

.500g of potassium hydrogen phthalate are titrated with .10 M sodium hydroxide.

a. How many moles of potassium hydrogen phthalate are being used in this titration?

b. when 20.00 ml of sodium hydroxide have been added, how many moles of sodium hydroxide will this be?

c. How many moles of potassium hydrogen phthalate will have reacted with this much sodium hydoxide?

d. How many moles of potassium hydrogen phthalate will be left unreacted?

e. How many moles of the salt will have been produced? Remembering that this salt will be completely ionized in the solution, how many moles of the anion will be present?

f. What will be the ration of the moles of anion to the moles of unreacted acid at this point?

g. How many ml of base would be required to neutralize half of this acid? (Referring to original titration).

I'm really stuck on d through g if someone could show the steps it would help tremendously.

Explanation / Answer

(a) Number of moles of potassium hydrogen pthalate = 0.500 / 204 = 0.00245 mol

(b) Number of moles of NaOH = 0.10 x 20 / 1000 = 0.002 mol

(c) 0.002 mol of potassium hydrogen pthalate reacts

(d) 0.00245 - 0.002 = 0.00045 mol pf potassium hydrogen pthalate is remained

(e) Number of moles of salt = number of moles of anion = 0.002 mol

(f) ratio = 0.002 / 0.00045 = 4.44

(g) Required base to neutralise half this acid = 12.25 mL

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