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Quick pH calculation -- Please verify
A 50.0 mL aqueous solution of sodium hydroxide has a pH of 12.50. If 36.00 ml of 0.0200 mol/L sulfuric acid is added to this sodium hydroxide solution, what will be the new pH of the resulting solution? Assume that the temperature stays constant at 25C, and that the volumes are perfectly additive.
50ml=0.05L
0.00158 mol OH- initially.
0.00072 mol H2SO4
Entire consumption of H2SO4(limiting reagent).
0.00086 mol OH- left.
0.00072 mol HSO4- left.
Entire consumption of HSO4-:
0.00014 mol OH- left.
Volume total=86 ml.
ph=11.21
Makes sense?
A 50.0 mL aqueous solution of sodium hydroxide has a pH of 12.50. If 36.00 ml of 0.0200 mol/L sulfuric acid is added to this sodium hydroxide solution, what will be the new pH of the resulting solution? Assume that the temperature stays constant at 25C, and that the volumes are perfectly additive.
50ml=0.05L
0.00158 mol OH- initially.
0.00072 mol H2SO4
Entire consumption of H2SO4(limiting reagent).
0.00086 mol OH- left.
0.00072 mol HSO4- left.
Entire consumption of HSO4-:
0.00014 mol OH- left.
Volume total=86 ml.
ph=11.21
Makes sense?