Author Topic: HELP part 2/2 (Read 563 times) Tweet Share

naved_s9994 · « **on:** August 29, 2009, 11:22:13 pm »

30ml of a substance has a pH of 3, 2.7L of water is added to it, what is the new pH?

i dont know what the substance is...?

TrueTears · « **Reply #1 on:** August 29, 2009, 11:25:23 pm »

You don't need to.

$pH = -log_{10}([H_3O^+])$

$[H_3O^+] = 10^{-3}$

$c_1V_1 = c_2V_2$

$10^{-3}(30 \times 10^{-3}) = c_2(2.7+30 \times 10^{-3})$

$c_2 = \frac{10^{-3}(30 \times 10^{-3})}{2.7+30 \times 10^{-3}}$

so $pH = -log_{10}(\frac{10^{-3}(30 \times 10^{-3})}{2.7+30 \times 10^{-3}})$

Search the forums now!

We have moved!

Author Topic: HELP part 2/2 (Read 563 times) Tweet Share

naved_s9994

HELP part 2/2

TrueTears

Re: HELP part 2/2

Recent Posts

User:
Password: