An HNO3 solution has a pH of 3.47. What volume of 0.015 M LiOH will be required to titrate 65.0 mL of the HNO3 solution to reach the equivalence point?
2007-04-20 19:20:33 · 3 answers · asked by sydney 1 in Science & Mathematics ➔ Chemistry
H+ con. in HNO3:
-log[H+]=3.47
[H+]=0.00034 mol dm-3
# of moles in 65.0ml of 0.0002 M HNO3= 0.00034/1000*65.0
=0.000022 mol
LiOH + HNO3 = LiNO3+ H2O
1 : 1
# of moles of 0.015M LiOH reacted=0.000022 mol
Required volume of 0.015M LiOH=0.000022/0.015
=0.00146 L
=1.46 mL
2007-04-20 19:38:59 · answer #1 · answered by Anonymous · 0⤊ 0⤋
HNO3 + LiOH -----> LiNO3 + H2O
It takes the same no of moles of LiOH as there were moles of HNO3 in the original solution. Since HNO3 is a strong acid, the concentration of H+ is the same as the concentration of HNO3. pH = -log[H+], so [H+] = 10^-3.47 = 3.39*10^-4M; 65mL has 0.065*3.39*10^-4 moles of H+ or 2.2*10^-5 moles H+. The moles of LiOH titrated will be V*0.015 (V in liters) so V*0.015 = 2.2*10^-5, V = 0.00147L or 1.47mL
2007-04-20 19:39:52 · answer #2 · answered by gp4rts 7 · 1⤊ 0⤋
Use the normality eqn.
2007-04-20 19:45:35 · answer #3 · answered by ag_iitkgp 7 · 0⤊ 0⤋