Question

HNO2 ( Ka=4.5 x 10^-4)

HCN (Ka= 4.9 x 10^-10)

CH3NH2 (Kb=4.4 x 10^-4)

HONH2 (Kb=1.1 x 10^-8)

Use this data to rank the following solutions in order of increasing pH. In other words, select a '1' next to the solution that will have the lowest pH, a '2' next to the solution that will have the next lowest pH, and so on..

RANK THESE ONES:

0.1 M NaBr

0.1 M CH3NH3Cl

0.1 M KNO2

0.1 M HONH3Br

Answer #1

We know that K_{a} x K_{b} =
1.0x10^{-14}

HNO_{2} ( K_{a}=4.5 x10^{-4})

HCN (K_{a}= 4.9 x10^{-10})

CH_{3}NH_{2} (K_{b}=4.4
x10^{-4}) ---> K_{a} = ( 1.0x10^{-14}
)/K_{b} = ( 1.0x10^{-14} ) / (4.4 x10^{-4})
= 2.27x10^{-11}

HONH_{2} (K_{b}=1.1 x10^{-8})
----> K_{a} = ( 1.0x10^{-14}
)/K_{b} = ( 1.0x10^{-14} ) / (1.1 x10^{-8})
= 9.09x10^{-7}

Also we know that As K_{a} increases [H^{+}]
concentration increases log [H^{+}] increases ---> - log
[H^{+}] decreases

= pH decreases

So high value of K_{a} has low pH ( The given
concentrations are same for all)

Order of K_{a} : HNO_{2} > HONH_{2}
> HCN > CH_{3}NH_{2}

Therefore the order of pH is : HNO_{2} < HONH_{2} < HCN < CH_{3}NH_{2}

So the order of given compounds is reversed(since the salts are taken)

KNO_{2} > HONH3Br > NaBr
> CH_{3}NH3Br

#1
Given that at 25.0 ∘C
Ka for HCN is 4.9×10−10 and
Kb for NH3 is 1.8×10−5,
calculate
Kb for CN− and
Ka for NH4+.
Enter the Kb value for CN− followed by the Ka
value for NH4+, separated by a comma, using two significant
figures.
#2 Find the pH of a 0.250 M solution of NaC2H3O2. (The
Ka value of HC2H3O2 is 1.80×10−5).

#1
Given that at 25.0 ∘C
Ka for HCN is 4.9×10−10 and
Kb for NH3 is 1.8×10−5,
calculate
Kb for CN− and
Ka for NH4+.
Enter the Kb value for CN− followed by the Ka
value for NH4+, separated by a comma, using two significant
figures.
#2 Find the pH of a 0.250 M solution of NaC2H3O2. (The
Ka value of HC2H3O2 is 1.80×10−5).

Calculate the pH of a 0.50 M solution of
methylamine(CH3NH2, Kb = 4.4 x
10-4.)
a.
1.83
b.
10.64
c.
12.17
d.
5.47
e.
8.53

Calculate the pH of a 0.22 M CH3NH3Br solution. Kb(CH3NH2) = 4.4
×10–4

In the titration of 40.00 mL of 0.500 M methyl amine, CH3NH2 (Kb
= 4.4 x 10–4) with 0.200 M HCl, calculate the pH of the resulting
solution when the titration is 1/4 of the way to the equivalence
point.

The Kb for methylamine, CH3NH2, at 25 C is 4.4 x 10-4 A. Write
the chemical equation for the equilibrium that corresponds to Kb.
B. By using the value of Kb, calculate ΔGo for the equilibrium in
part A. C. What is the value of ΔG at equilibrium? D. What is the
value of ΔG when [H+] = 1.6 ×10-8 M, [CH3NH3+] = 5.8 ×10-4 M, and
[CH3NH2] = 0.130 M?

Find the pH of 0.180 M NaCN solution. For HCN,
Ka=4.9⋅10^−10

1.ka for HF is 6.8x10^-4. calculate the kb for its
conjugate base, the flouride ion, F-
2. which of the following salts dissovled in water
will form basic solutions?
NH4Cl? Cu(No3)2? NaCN? LiF?
3. what is the concentration (in M) of hydronium ions
[H3O+] or [H+] in a solution at 25°C with pH=4.282
4. what is the pOH of a solution with a pH of 3.7?
5. the kb of methylamine is 4.4 x 10^-4. calculate tje
pH of a...

Calculate the pH of a 0.375 M aqueous solution
of nitrous acid (HNO2,
Ka = 4.5×10-4) and the
equilibrium concentrations of the weak acid and its conjugate
base.
pH
=
[HNO2 ]equilibrium
=
M
[NO2-
]equilibrium
=
M

Calculate the percent dissociation of HNO2
(Ka=4.5×10−4) in 0.050 M HNO2.

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