In: Chemistry
Calculate the pH, pOH, [H3O+], and [OH-] of the following
solutions:
(a) 0.001 M HNO3,
(b) 0.3 M Ca(OH)2;
(c) 3.0 M HNO3,
(d) 6.0 M NaOH,
(e) 0.05 M HBr
(a) 0.001 M HNO3
HNO3 is strong acid
HNO3 + H2O--------> H3O+ + NO3-
0.001M 0.001M
[OH-] = Kw/[H3O+]
= 1*10-14/0.001 = 10-11 M
PH = -log[H3O+]
= -log0.001 = 3
(b) 0.3 M Ca(OH)2
Ca(OH)2 ---------> Ca+2 + 2OH-
0.3M 2*0.3 = 0.6M
[OH-] = 0.6M
[H3O+] = Kw/[OH-]
= 1*10-14/0.6 = 1.67*10-14 M
PH = -log1.67*10-14 =
13.7772
(c) 3.0 M HNO3
HNO3 + H2O---------> H3O+ + NO3-
3M 3M
[H3O+] = 3M
[OH-] = Kw/[H3O+]
= 1*10-14/3 = 3.4*10-15 M
PH = -log[H3O+]
= -log3 = -0.4471
(d) 6.0 M NaOH,
NaOH -------> Na+ + OH-
6M 6M
[OH-] = 6M
[H3O+] = Kw/[OH-]
= 1*10-14 /6 = 1.7*10-15 M
PH = -log[H3O+]
= -log1.7*10-15
= 14.7695
(e) 0.05 M HBr
HBr ----------> H+ + Br-
0.05M 0.05M
[H+] = 0.05M
[OH-] = Kw/[H3O+]
= 1*10-14 /0.05 = 2*10-12 M
PH = -log[H3O+]
= -log0.05 = 1.3010