Chemistry Homework Please Help

Question

Please if someone could please help me figure this out I would be so greatful I have been stuck on this problem for a few days and I cant get it I have to show all work for my answer. 

 

300 mL of 0.06 M HNO3 solution are poured into 34 liters of water. 450 mL of 3.2 x 10-4 M H2SO4 solution are also poured into this water. If 250 mL of this final solution were transferred to a second container, how many liters of water would be needed to dilute the solution to a pH of 5.0 ?

Answer 1

300 mL of 0.06 M HNO3 solution are poured into 34 liters of water

(300mL)(0.06 mol/1000 mL HNO3) = .018 mol of H⁺

450 mL of 3.2 x 10-4 M H2SO4 solution are also poured into this water

(450 mL)(3.2 x 10^-4 mol/1000mL H2SO4) = 1.44 x 10^-4 mol H2SO4, but twice as much H⁺ because there are two H for each H2SO4

So, 2 x 1.44 x 10^-4 mol H2SO4 = 2.88 x 10^-4 H⁺

Sum these together:

2.88 x 10^-4 H⁺  +  .018 mol of H⁺ = .018288 mol H⁺

This amount of H⁺ is in 34L + .300L + .450 L

That's a total volume of 34.750L

.018288 mol H⁺/34.75 L = 5.26 x 10^-4 Molar of H⁺

Take the negative log of that and get a pH of 3.28.  But, want a pH of 5.  A pH of 5 means that the concentration of H⁺ is 1 x 10^-5

So, .018288 mol H⁺/xL = 1 x 10^-5

x = 1828.8 L

1828.8 L - 34.75 L = 1794.05 L of water must be added.

 

I hope someone like Ishmar or JR, who are better chemists than I can check this solution.