Calculations (1 Viewer)

[toknblackguy]

how do u calcualte the neutralisation questions
eg, 15ml 0.02M nitric acid is added to 10ml 0.01M barium hydroxide. calculate pH of solution (assume complete dissociation of the acid and base)

thanks

 

[spice girl]

Originally posted by toknblackguy
how do u calcualte the neutralisation questions
eg, 15ml 0.02M nitric acid is added to 10ml 0.01M barium hydroxide. calculate pH of solution (assume complete dissociation of the acid and base)

thanks

using n=cV
have n(H+) = 0.015 * 0.02 = 3 * 10^-4 mol
n(OH-) = 0.010 * 0.01 * 2 = 2 * 10^-4 mol
have excess 1 * 10^-4 mol of H+ in total of 25mL
this equates to [H+] = 4 * 10^-3 M
pH = -log(H+) = wotever,...

 

[toknblackguy]

u lost me at "have excess 1 times..."

 

[Frigid]

lemme try...

N(HNO3) = C x V = 0.015 x 0.02 = 3 x 10^-4 moles

N(Ba(OH)2) = C x V = 0.01 x 0.01 = 1 x 10^-4 moles

Now the neutralisation equation is:

2HNO3(aq) + Ba(OH)2(aq) ---> 2H2O(l) + Ba(NO3)2(aq)

That means the stoichimetric ratio N(HNO3) : N(Ba(OH)2) = 2 : 1

Therefore, since we have 1 x 10^-4 moles of Ba(OH)2, it will neutralise with 2 x 10^-4 moles of HNO3.

Now we look at the products - nope, Ba(NO3)2 is not an acidic or basic salt. good.

But we have 3 x 10^-4 moles of HNO3 and only 2 x 10^-4 moles is used. Therefore 1 x 10^-4 moles of HNO3 remains in solution.

Assuming complete dissociation, HNO3 -> H+ + NO3-

Therefore 1 x 10^-4 moles of H+ formed. But this is not completely true, since to calculate pH we need all of this in moles per litre.

1 x 10^-4 divided by (0.015 + 0.01)* = 4 x 10^-3 moles/litre

Now we can chuck into negative log eqn.

pH = - log[H+] = 2.398 (3dp)

corrections graciously accepted .

NB: * (0.015 + 0.01) is the total amount of solution.

 

[toknblackguy]

woah
thanks heaps frigid!
and spice girl of course
i guess i can safely post like crazy here

 

[mon_mon]

This is a titration, i assume? because thats my weakest point, but i found that surprisingly easy to understand. thanks fridgid.