Question

how do u find the resulting ph when given the volume and a ph for a acid and a base

Answer 1

Do you have a sample question? I can walk you through it.

Answer 2

yes i do

Answer 3

its what is the resulting ph when 250ml of HCL solution with ph 1.50 is mixed with 250ml of a NaOH solution with ph of 13.50

Answer 4

I also have another question after tht.

Answer 5

Okay, first step, determine the concentrations & moles of HCl and NaOH. Do you know how to do that?

Answer 6

so to find the concentration you use the formula c=n/v

Answer 7

and n= moles over molar mass

Answer 8

Well, to find the concentrations, you need to know the pH's and pOH's of each solutions. Since HCl and NaOH are both strong acids, they both dissociate. You can treat the NaOH as just OH- and the HCl as H-.

[H+] = 10^(-pH)
[HCl] = [H+] = 10^(-1.5) = 3.16 x 10^-2 M

[OH-] = 10^(-pOH)
pH + pOH = 14, pH = 13.5 ---> pOH = 0.5
[NaOH] = [OH-] = 10^(-0.5) = 3.16 x 10^-1 M

Now, you can find the moles using the concentration.\[n = MV\]\[n_{NaOH} = 3.16 \times 10^{-1} * 0.250 L = 7.91 \times 10^{-2} mol\]\[n_{HCl} = 3.16 \times 10^{-2} * 0.250 L = 7.91 \times 10^{-3} mol\]

Answer 9

my other question is when 20.0ml og 0.100mol/L of HCl is mixed with a solution of NaOH with a pOH of 1.05 the resulting solution has a total volume of 50.0mL. wht is the ph of this mixture?

Answer 10

\[HCl + NaOH \rightarrow NaCl + H_2O\]So in the reaction that takes place when the 2 solutions are mixed, the acid and base neutralize each other to form water and a salt. There is more NaOH, so it is going to get 'rid of all the HCl.
7.91 x 10^-2 - 7.91 x 10^-3 = 7.11 x 10^-2 mol NaOH

Answer 11

If you understand the first, you can do the second.

Answer 12

yes.. i do
thanks alot for ur help.
plus do u have any sites in mind tht would help me with this stuff

Answer 13

like any sources?

Answer 14

Now, that NaOH is a strong base that remains, all the HCl has been neutralized.
\[NaOH \rightarrow Na^+ + OH^-\]Now all of the NaOH is going to become Na+ and OH-. So your moles of OH- is 7.11 x 10^-2. The new concentration of the solution is 500 mL = 0.5 L. The concentration of OH- is\[n = MV = 0.5 L \times 7.11 \times 10^{-2} = 3.56 \times 10^{-2}\]Now, \[pOH = - \log_{10} [OH^-] = - \log_{10} 3.56 \times 10^{-2} = 1.45 \]pOH + pH = 14
1.45 + pH = 14
pH = 12.55

So the pH of the new solution is 12.55.

Answer 15

A good website I know is chem team. It has concepts, examples, problems, detailed solutions, etc.
http://www.chemteam.info/ChemTeamIndex.html