chemistry plz?

Question

1.equal volumes of 2.0 mol/L of NaoH solution and the 0.2 HCL mol/L are mixed .What's the pH value?
2.ifr equal volumes of 0.2mol/L of HCLand 2.0 mol/L of NaAc

Answer 1

1) NaOH and HCl are a strong acid and a strong base. The resulting solution (0.2 mol/L each) will form a 0.2 mol/L NaCl solution which will have the same pH as just water (pH =7.00)

2) NaAc is Sodium acetate. It is considered a relatively strong base, but not like NaOH. It will react, in part to give

HCl + NaAc --> NaCl + HAc

HAc is Acetic acid. What you end up with is a buffered solution of about 0.2 mol/L Acetic acid and 1.8 mol/L Sodium acetate. This will have a pH different than 7.00.

To calculate this pH:

[H+] = Ka of HAc * [HAc]/[Ac-]

= (1.7 * 10^-5) * (0.2/1.8) = 0.189 * 10^-5

just convert this to pH.

Answer 2

NaOH and HCl are a strong acid and a strong base. The resulting solution (0.2 mol/L each) will form a 0.2 mol/L NaCl solution which will have the same pH as just water (pH =7.00)

2) NaAc is Sodium acetate. It is considered a relatively strong base, but not like NaOH. It will react, in part to give

HCl + NaAc --> NaCl + HAc

HAc is Acetic acid. What you end up with is a buffered solution of about 0.2 mol/L Acetic acid and 1.8 mol/L Sodium acetate. This will have a pH different than 7.00.

To calculate this pH:

[H+] = Ka of HAc * [HAc]/[Ac-]

= (1.7 * 10^-5) * (0.2/1.8) = 0.189 * 10^-5

just convert this to pH.
it as simple as that well got to go

Answer 3

1) Find out which of the two are inexcess. (find number of mole of each solutions by n=CV).Then, use the nos of mole of the inexcess solution to find the ph (pH = log10(CONC of the solution)) From the looks of it, the naoh is inexcess, so the ph should be above 7.
2) (note i think the NaAc's supposed to be NaOH), so use same method as above

Answer 4

Eh isso mesmo mas como faço pra enteder isso????

Answer 5

oh yeah!!