Calculate the pH of a 0.0152 M aqueous solution
of nitrous acid (HNO2,
Ka = 4.6×10-4) and the
equilibrium concentrations of the weak acid and its conjugate
base.
pH
=
[HNO2]equilibrium
=
M
[NO2-
]equilibrium
=
M
Determine the pH of each solution.
a. 0.155 M HNO2 (for HNO2, Ka=4.6×10^−4)
b. 0.0210 M KOH
c. 0.240 M CH3NH3I (for CH3NH2, Kb=4.4×10^−4)
d. 0.324 M KC6H5O (for HC6H5O, Ka=1.3×10^−10)
what is thE ph for
A. 0.021 M hno2 at 25 degrees with ka of 4.3*10^-4
B. .15 M nh2ch5 at 25 with kb of 5.4*10^-5
C. .45 m k2so4 at 25 with kb of 8.33*10^-13
HNO2 (aq) + NaOH (aq) → NaNO2 (aq) +
H2O
(l)
Ka of HNO2 = 4.5 x 10-4
H2O (l) + NO2- (aq) ↔
HNO2 (aq) + OH- (aq)
Kb of NO2- = 2.2 x
10-11
Exactly 100 mL of 0.10 M nitrous acid (HNO2) are
titrated with a 0.10 M NaOH solution. Calculate the pH for:
A:The initial solution
B:At the half-equivalence point
C:The point at which 80 mL of the base has been added
D:The equivalence point...
What is the pH of a 0.25 M solution of KHCOO? Ka (HCOOH) = 1.8 *
10-4
Please show all the steps because I'm doing something wrong and
keep getting 2.17. Thank You
calculate the ph of 1.0 l of a buffer that is .01 m of HNO2 and
.15 m NaO2. what is the ph of the same buffer and after the additon
of 1.0 l of 12m hcl (pka of HNO2 is 3.4)