A student uses a neutralization reaction to determine the sulfuric acid (H_2SO_4

Question

A student uses a neutralization reaction to determine the sulfuric acid (H_2SO_4) concentration in a H_2SO_4 solution. The reaction of H_2SO_4 with NaOH is shown in Equation 13. H_2SO_4(aq) + 2 NaOH(aq) rightarrow Na_2SO_4(aq) + 2 H_2O(l) The student dilutes 20 mL of the H_2SO_4 solution to 100 mL. A 25.00-mL aliquot of the diluted H_2SO_4 solution requires 39.78 mL of 0.1073M NaOH to reach a phenolphthelein end point. (1) Calculate the molarity of the diluted H_2SO_4 solution. (2) Calculate the molarity of the original H_2SO_4 solution.

Explanation / Answer

NaOH moles = M x V = 0.1073 x (39.78/1000) = 0.0042684

H2SO4 moles in 25 ml aliquot = NaOH moles/2 = 0.004284/2 = 0.0021342

( since H2SO4 and NaOH reacts in 1:2 ratio)

Now 25 ml solution has 0.0021342 moles H2SO4 , then 100 ml of H2SO4 solution has

H2SO4 moles = ( 100/25) x 0.0021342 = 0.0085368

1) Molarity of diluted solution = moles/ vol in L of H2SO4 = 0.0085368 /0.1 = 0.08537 M  

2) Initial vol is 20 ml , hence Molarity = moles/ vol 0.0085368 /0.02 = 0.427 M

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