Acid – Base Equilibria and Rates of Reaction PLEASE HELP
1) Calculate the pH of the following solutions at 298K:
(a) 0.15 mol dm-3 HCl
(b) 0.10 mol dm-3 H2SO4
(c) 0.10 mol dm-3 benzoic acid (Ka = 6.4 x 10-5 mol dm-3)
(d) 0.05 mol dm-3 KOH (Kw(298) = 1 x 10-14 mol2 dm-6)
(a) 0.15 mol dm-3 HCl
(b) 0.10 mol dm-3 H2SO4
(c) 0.10 mol dm-3 benzoic acid (Ka = 6.4 x 10-5 mol dm-3)
(d) 0.05 mol dm-3 KOH (Kw(298) = 1 x 10-14 mol2 dm-6)
(edited 9 years ago)
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What have you got so far?
Original post by EierVonSatan
What have you got so far?
Nothing I don't know where to start....
Original post by soni88
Nothing I don't know where to start....
Okay lets try the first one, the formula for pH is:
Since HCl is a strong acid, it is fully dissociated in water. That help?
a) Is a strong acid so it fully dissociates, therefore concentration of H+ is equal to the concentration of HCl. Remember pH = -log[H+]
b) Is a diprotic strong acid. So one mole of H2SO4 fully dissociates into two moles of H+ ions. Therefore -log[0.2] will give you its pH
c) Benzoic acid is a weak acid. So it's an equilibria system. Ka indicates the position of equilibrium (so therefore the extent of dissociation). You need to write the Ka equation for Benzoic acid. Remember that we assume that the concentration of benzoic acid at eqm is equal to its initial conc because the extent of dissociation is so little. Also we assume that [H+]=[A-] which we simplify to [H+]2 because the degree of dissociation of water is negligible
d) KOH is a strong base therefore it fully dissociates into K+ and OH-. It's a 1:1 ratio therefore [OH-]=[KOH initial]. Kw = [OH-][H+], rearrange to get [H+] and then -log it to get pH.
Hope this helped? Let me know if anything needs clarifying.
b) Is a diprotic strong acid. So one mole of H2SO4 fully dissociates into two moles of H+ ions. Therefore -log[0.2] will give you its pH
c) Benzoic acid is a weak acid. So it's an equilibria system. Ka indicates the position of equilibrium (so therefore the extent of dissociation). You need to write the Ka equation for Benzoic acid. Remember that we assume that the concentration of benzoic acid at eqm is equal to its initial conc because the extent of dissociation is so little. Also we assume that [H+]=[A-] which we simplify to [H+]2 because the degree of dissociation of water is negligible
d) KOH is a strong base therefore it fully dissociates into K+ and OH-. It's a 1:1 ratio therefore [OH-]=[KOH initial]. Kw = [OH-][H+], rearrange to get [H+] and then -log it to get pH.
Hope this helped? Let me know if anything needs clarifying.
Original post by soni88
how do i do them on a calculator?
Again, for the first one the concentration of the acid is the same as the concentration of H+. Using the formula above then put into the calculator:
[-] [log] [0] [.] [1] [5] [=]
tell us what you get
Original post by soni88
0.82390874
Yeah, but we normally round answers to 3 significant figures
0.824
Original post by soni88
so after this what do i do?
Take a look at LeaX's post above and try part b
It's the same formula, and similar calculator buttons.
Original post by soni88
is the answear for b 0.698
Yes, 0.699 to 3 significant figures.
Original post by soni88
does that mean the pH is round 7
No the pH is 0.699. That's very acidic, pH 7 is neutral.
Is anyone going over this for the chem empa (A2)? has anyone done it yet?
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