0.300g of solid NaOH was added to 1.00L of 5.00*10^-3 mol/L HNO3
Assuming no volume change, what is the pH of final solution
ans: 11.4
what i did was:
-show that HNO3 is limiting reagent with moles (5*10^-3)
-n(H20) = (5*10^-3) since 1 mol of HNO3 produce 1 mole of H20
-find n(H+)= n(H20)
-[H+]= n/v
- pH=-log(H+)
Hey Amanda:
Good question. Ok so for me firstly what I would have done is to establish a balanced neutralisation equation.
NaOH (s) + HNO3 (aq) --> NaNO3 (aq) + H2O (l)
So clearly with this equation no further balancing is required. Now we need to work out the moles of HNO3 and moles of NaOH to determine the limiting reagent.
n of NaOH = 0.3/(22.9 + 16 + 1.008) = 0.0075173 moles
n of HNO3 = CV = 5 x 10^-3 x 1 = 0.005 moles
Ok so quite clearly since the molar ratio in the balanced neutralisation equation is 1:1:1:1, HNO3 is the limiting reagent, as you have correctly identified, well done!
Now theres an extra step here we need to be careful of, because the H+ ions and OH- ions that are dissociated from HNO3 and NaOH would neutralise one another and become H2O, we need to calculate the amount of moles of NaOH that is in excess of:
Excess n(NaOH) = 0.0075173 - 0.005 = 0.0025173 moles
Assuming theres no volume change, the concentration of NaOH would be C = 0.0025173/1 = 0.0025173 mol/L
So now we know that there are 0.0025173 moles per litre of OH- ions in the solution. In order to figure out the concentration of H+ ions and the pH values, we need to use the water constant (Kw = 1.0 x 10^-14).
Kw = [H+] [OH-]
[H+] = 1.0x10^-14/0.0025173 = 3.97251 x 10^-12
Now we can calculate the pH:
pH = -log(3.97251 x 10^-12)
Hence pH = 11.4
Hope you understood my reasoning there amanda, a very good question indeed! If you have any further queries please dont hesitate to ask!
Best Regards
Happy Physics Land