A student titrated 50mL of an H2SO4 solution with .2M NaOH using a phenolphthale
ID: 823625 • Letter: A
Question
A student titrated 50mL of an H2SO4 solution with .2M NaOH using a phenolphthalein indicator. The student found that it took 50mL of NaOH to reach the equivalence point. Using the following equation: M1V1=M2V2 the student was able to determine the concentration of the H2SO4 solution to also be .2M. Is this the correct concentration of the unknown H2SO4 solution? If not, what did the student do incorrectly?
What is the ratio between the average number of drops of sodium hydroxide solution required to neutralize 2.0mL of sulfuric acid and 2.0mL of hydrochloric acid?
Explanation / Answer
(1) The student is incorrect.
H2SO4 + 2 NaOH => Na2SO4 + 2 H2O
NaOH: M1 = 0.2, V1 = 50 mL
H2SO4: M2 = ?, V2 = 50 mL
From stoichiometry of reaction:
Moles of NaOH = M1V1 = 2 x moles of H2SO4 = 2 x M2V2
0.20 x 50 = 2 x M2 x 50
Concentration of H2SO4 = M2 = 0.1 M
(2) H2SO4 + 2 NaOH => Na2SO4 + 2 H2O
HCl + NaOH => NaCl + H2O
Drops of NaOH for H2SO4/drops of NaOH for HCl
= moles of NaOH for H2SO4/moles of NaOH for HCl
= 2/1 = 2
Thus the ratio is 2 : 1
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