Helpp!!!

[aero135]

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20 mL of 0.08 mol L–1 HCl is mixed with 30 mL of 0.05 mol L–1 NaOH.
What is the pH of the resultant solution?
(A) 1.1
(B) 2.7
(C) 4.0
(D) 7.0

how do u do this please show working

also how do u do question 12 from the 2008 HSC paper


Samples of water were collected from a river at four different sites: forest, mine, town and estuary. Ocean Estuary sampling site Mine sampling site Copper mine discharge Town sampling site Town sewage treatment plant discharge Forest catchment Forest sampling site

The results of various analyses of the water samples are shown. Site 1	Site 2	Site 3	Site 4
pH Total Dissolved Solids (mg/L) Biochemical Oxygen Demand (mg/L) E. coli (CFU/100 mL)				6.8 305 32 18	6.8 85 2 2	7.8 7600 2 2	5.9 290 3 2

cheers

[Fortify]

First one:

n(HCl) = c * v = 0.08 * 20 = 1.6mol

n(NaOH) = c * v = 0.05 * 30 = 1.5mol

Therefore HCl is in excess by 0.1mol.

Now,

c = n / v = 0.1 / 50 = 2*10^-3mol/L

pH = - log [H+] = - log [2*10^-3] = 2.69 = 2.7 pH

[Fortify]

And for Q12. I think it's A.

[study-freak]

Originally Posted by Fortify

First one:

n(HCl) = c * v = 0.08 * 20 = 1.6mol

n(NaOH) = c * v = 0.05 * 30 = 1.5mol

Therefore HCl is in excess by 0.1mol.

Now,

c = n / v = 0.1 / 50 = 2*10^-3mol/L

pH = - log [H+] = - log [2*10^-3] = 2.69 = 2.7 pH

Wrong working although the answer is correct.

n(HCl) = c * v = 0.08 * 0.02 = 0.0016mol

n(NaOH) = c * v = 0.05 * 0.03 = 0.0015mol

Therefore HCl is in excess by 0.0001mol. (and hence H3O+ (or H+) ions by the same amount since HCl(aq) + NaOH(aq) -> H2O(l) + NaCl(aq) and HCl(aq) + H2O(l) ->H3O+(aq) + Cl-(aq) )

Now,

c(H+) = n / v = 0.0001 / 0.05 = 2*10^-3mol/L

pH = - log [H+] = - log [2*10^-3] ~ 2.699 ~ 2.7

Hence (B)

[Fortify]

It does not matter. As long as my measurements are kept constant or the same (ie: mL)

[study-freak]

Originally Posted by Fortify

It does not matter. As long as my measurements are kept constant or the same (ie: mL)

It does matter because your units are incorrect then. (not the answer but in your working)

[Fortify]

Originally Posted by Fortify

First one:

n(HCl) = c * v = 0.08 * 20 = 1.6mol

n(NaOH) = c * v = 0.05 * 30 = 1.5mol

Therefore HCl is in excess by 0.1mol.

Now,

c = n / v = 0.1 / 50 = 2*10^-3mol/mL

pH = - log [H+] = - log [2*10^-3] = 2.69 = 2.7 pH

happy?

chillax bro.

[study-freak]

Originally Posted by Fortify

happy?

chillax bro.

Your units for n(HCl) and n(NaOH) shouldn't be mol.
And 2*10^-3mol/L is correct, not mol/mL.

[Fortify]

you prosu.