year 11 ph question

[Kaille]

heyo,
having trouble...

What volume of water is required to dilute 30 ml of a nitric acid solution of ph3 to ph5?

any help much appreciated

[REBORN]

Do you have answers? I'm in y11 too but I've covered this....and think I got it? -.-

My anwer is add 2970mL (2.97L).....DO NOT take it as correct until confirmation is given by some beast (ie luken93/thushan)

[Kaille]

you are correcto! how did you do it?

[|ll|lll|]

n = m/M
n = cV
3 = -log_10 [HNO3]
c(HNO3)= 0.001
n(HNO3) = 0.001 x 0.03 = 3 x 10^-5

We want c = 10^-5
V = 3 x 10^-5 / 10^-5 = 3L

Hence 3000-30 = 2970mL

[HenryP]

The pH system uses a base of 10. What this means is that to go from one whole number to another you have to multiply/divide by 10.
Taking the example you have, if you have 30ml of nitric acid which is at pH 3. To get a solution of pH 4 you need to dilute it too 300ml. From there you need to dilute by another factor of 10 to get it to pH 5 which will bring the total volume up to 3000ml. Therefore adding 2970 ml, you will get a solution of pH 5

[REBORN]

I did it as | | | | girl thingo did it.

Henry's way is 10x quicker though.

[Kaille]

thanks for the help guys

[luken93]

Or easier, C1V1 = C2V2
Where C1 = 1x10^-3   <= Since c(H+) = 10^-pH
V1 = 0.03
C2 = 1x10^-5
V2 = ?

10^-3 x 0.03 = 10^-5 x V2
V2 = 3
Since that's final, 3 - 0.03 = 2.97 = 2970mL