This has a very good explaination. Go through his examples, especially at the end.
On c, find the OH concentration, then the pOH, then you know pH+pOH=14 and you can solve for pH from that.
Calculate the pH of each of the following solutions:
a) .35 M hydrochloric acid, HCl
b) .35 M acetic acid, HC2H3O2
c) .35 M sodium hydroxide, NaOH
Can someone please explain to me how to calculate the pH in a way that is easy to understand? I am confused. Could you please do one of the above questions as an example? Thank you so much!
3 answers
a) 0.35M HCl (monoprotic strong acid) produces a 0.35M [H+] solution.
pH = -log(0.35)
c) 0.35M NaOH (strong base) acid produces a 0.35M [OH-] solution.
pOH = -log(0.35)
pH = 14-pOH
b) 0.35M HC2H3O2 (WEAK ACID) does not dissociate completely. Its Ka = 1.8x10^-5
[H+] = sqrt[(Ka)(0.35)]
pH = -log[H+]
pH = -log(0.35)
c) 0.35M NaOH (strong base) acid produces a 0.35M [OH-] solution.
pOH = -log(0.35)
pH = 14-pOH
b) 0.35M HC2H3O2 (WEAK ACID) does not dissociate completely. Its Ka = 1.8x10^-5
[H+] = sqrt[(Ka)(0.35)]
pH = -log[H+]
True
1 answer