Acid question :) (1 Viewer)

[dwayne95]

Bit confused by this one - help/method would be appreciated.

"A 0.1 mol/L HCl solution has a pH of 1.0.

What volume of water must be added to 90mL of this solution to obtain a final pH pf 2.0?

a) 10mL
b) 180 mL
c) 810 mL
d) 900 mL"

 

[dwayne95]

PS - sorry if this is in the wrong section.
PPS - am I right in thinking C is right? Maybe??

 

[umm what]

i hate this question
thats in one of the past papers ...i think HSC 2006/7

PS - sorry if this is in the wrong section.
PPS - am I right in thinking C is right? Maybe??

 

[nightweaver066]

Since the pH increases by 1, the concentration of hydronium ions decreases by 10 fold meaning volume increases by 10 fold.

So if it was initially 90mL, you want it to be 900mL, so you add 810mL of water.

 

[umm what]

Since the pH increases by 1, the concentration of hydronium ions decreases by 10 fold meaning volume increases by 10 fold.

So if it was initially 90mL, you want it to be 900mL, so you add 810mL of water.

how do we know this?

 

[jnney]

how do we know this?

Well, for the pH to be 2, the solution must be 10 times less acidic as it was originally.

 

[Ms106]

You can use c1v1=c2v2
Where c1 = 0.1mol/L
v1 = 90
c2 = 0.01
v2 = ?

Therefore you get v2 = 900mL
And you already have 90mL, so it's 900mL-90mL = 810mL

 

[VJ30]

c1v1=c2v2

c1=0.1 M
v1=0.9 L
c2=0.01 M
v2=?

Doing the calculations v2=.9L or 900mL....But u already have 90mL so the answer is 810mL !!

 

[Merase]

Since it hasn't been formally mentioned, the preliminary formula required for this Q is:

[H^+] = 10^(-pH)
(derived from the original pH = - log[H^+])

#potentially redundant chem post

 

[trolly]

its 10 because you carry the zero form the other number and then remove 0 off the end gives you 10ml. Btw im a physisist i know my chem.