Back Titration Q Help!

[gorillafunk]

Hello fellow Chemists, need some help with this question
A 50 mL volume of 0.10 M nitric acid is mixed with 60 mL of 0.10 M calcium hydroxide solution. What volume of 0.050 M sulfuric acid is required to neutralise the mixture?






Answer is 70 mL

Thanks :uglystupid2:

[totaled]

2HNO3 + Ca(OH)2 -> Ca(NO3)2 + 2H2O
thus, n(HNO3) = cv = 0.1 x 0.05 = 0.005mol
n[Ca(OH)2] = cv = 0.1 x 0.06 = 0.006mol
you can tell that we have excess Ca(OH)2, 0.006 - 0.0025 = 0.0035mol in excess
therefore, we need to write the eqn to figure out how to neutralise this excess base, with sulfuric acid

Ca(OH)2 + H2SO4 -> CaSo4 + 2H2O
n[Ca(OH)2] = excess from previous reaction = 0.0035mol
therefore as it is a 1 to 1 ratio, we NEED 0.0035mol of H2SO4 to neutralise the base.
as n=cv, v(H2SO4) = n/c = 0.0035/0.050 = 0.07L = 70mL as required