Question
What is the pH of a solution when 500ml of 0.2M NaOH and 500ml of 0.2M H2SO4 are mixed? Pls help.
Answers
2NaOH + H2SO4 ==> 2H2O + Na2SO4
You have 0.5L x 0.2M = 0.1 mol NaOH
You have 0.5L x 0.2M = 0.1 mol H2SO4.
Which is the limiting reagent? It must be NaOH; i.e., 0.1 mol H2SO4 will take 0.2 mol NaOH for complete neutralization and you don't have that much; therefore, the NaOH will neutralize 1/2 the H2SO4 and leave NaHSO4. The pH will be determined by the ionization of HSO4^-. It was 0.2, half is gone to make it 0.1 and that has been diluted by 1/2 to make the concn = 0.05.
..........HSO4^- ==> H^+ + SO4^2-
initial...0.05M......0.......0
change.....-x.........x.......x
equil....0.05-x.......x.......x
Substitute from the ICE chart into K2 for H2SO4 and solve for x, then convert to pH.
You have 0.5L x 0.2M = 0.1 mol NaOH
You have 0.5L x 0.2M = 0.1 mol H2SO4.
Which is the limiting reagent? It must be NaOH; i.e., 0.1 mol H2SO4 will take 0.2 mol NaOH for complete neutralization and you don't have that much; therefore, the NaOH will neutralize 1/2 the H2SO4 and leave NaHSO4. The pH will be determined by the ionization of HSO4^-. It was 0.2, half is gone to make it 0.1 and that has been diluted by 1/2 to make the concn = 0.05.
..........HSO4^- ==> H^+ + SO4^2-
initial...0.05M......0.......0
change.....-x.........x.......x
equil....0.05-x.......x.......x
Substitute from the ICE chart into K2 for H2SO4 and solve for x, then convert to pH.
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