What's an IRF?
2NaOH + H2SO4 = Na2SO4 + 2H2O
Convert pH of NaOH to pOH and calculate (OH^-) and (NaOH). Then mols NaOH = M x L = ?
mols H2SO4 = M x L = ?
Subtract to see which is in excess, calculate (H^+) or (OH^-) as it applies and convert to pH.
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 25 C, and that the volumes are perfectly additive.
I dont understand how what to do? Like do i do IRF or ICE tables or what? Please write a step by step. Thanks
1 answer
1 answer