1.A sample of potassium hydrogen phthalate (KHC8H4O4), which is an acid, with a
ID: 694033 • Letter: 1
Question
1.A sample of potassium hydrogen phthalate (KHC8H4O4), which is an acid, with a mass of 0.4749 g is dissolved in water and used to standardize a solution of strontium hydroxide. An endpoint is reached when 38.96 mL of strontium hydroxide has been added. Determine the molarity (in mol/L) of the strontium hydroxide solution. Report your answer to four significant figures.
2.The standardized strontium hydroxide solution is then used to titrate 27.22 mL of a solution of formic acid (HCOOH). An endpoint is reached when 15.97 mL of strontium hydroxide has been added. Determine the molarity (in mol/L) of the formic acid solution. Report your answer to four significant figures.
Explanation / Answer
1) 2KHC8H4O4 + Sr(OH)2 -------> Sr(KC8H4O4)2 + 2H2O
soichiometrically, 1mole of Sr(OH)2 react with 2mole of KHC8H4O4
mass of KHP = 0.4749g
molar mass of KHP = 204.22g/mol
No of mole of KHP = 0.4749g/204.22(g/mol) = 0.002325mol
0.002325 mole of KHP react with 0.0011625mole of Sr(OH)2
Volume of Sr(OH)2 solution = 38.96ml
Molarity of Sr(OH)2 = (0.0011625mol/38.96ml)×1000ml = 0.02984M
2) Sr(OH)2 + 2HCOOH ---------> Sr(HCOO-)2 + 2H2O
stoichiometrically, 1mole of Sr(OH2 react with 2mole of HCOOH
No of mole of Sr(OH)2 consumed = (0.02984mol/1000ml)×15.97ml = 0.0004765mol
0.0004765mol of Sr(OH)2 react with 0.000953mol of HCOOH
Volume of HCOOH solution = 27.22ml
Molarity of HCOOH = (0.000953mol/27.22ml)×1000ml = 0.0350M
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