ATAR Notes: Forum

VCE Stuff => VCE Science => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Chemistry => Topic started by: naved_s9994 on August 29, 2009, 11:22:13 pm

Title: HELP part 2/2
Post by: 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...? :-[
Title: Re: HELP part 2/2
Post by: TrueTears 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 $pH = -log_{10}(\frac{10^{-3}(30 \times 10^{-3})}{2.7+30 \times 10^{-3}})$