Question

**#1**

Given that at 25.0 ∘C

*K*a for HCN is 4.9×10−10 and

*K*b for NH3 is 1.8×10−5,

calculate

*K*b for CN− and

*K*a for NH4+.

Enter the *K*b value for CN− followed by the *K*a
value for NH4+, separated by a comma, using two significant
figures.

#2 Find the pH of a 0.250 *M* solution of NaC2H3O2. (The
*K*a value of HC2H3O2 is 1.80×10^{−5}).

Answer #1

**1.Since we know that at standard conditions at
25 ^{o}C**

**K _{a}*K_{b}=K_{w}=10^{-14}**

**so if K _{a} for HCN=4.9 *
10^{-10}**

**thus K _{b} for
CN^{-}=10^{-14}/4.9 *
10^{-10}=2*10^{-5}**

**Similarly, if K _{b} form
NH_{3}=1.8*10^{-5}**

**So,K _{a} for
NH_{4}^{+}=10^{-14}/1.8*10^{-5}=5.56*10^{-10}**

**2.
HC _{2}H_{3}O_{2}+H_{2}O-->H_{3}O^{+}+C_{2}H_{3}O_{2}^{-},
K_{a}=1.8*10^{-5}**

**But when
Na****C _{2}H_{3}O_{2} is added
in water,it acts as base**

**C _{2}H_{3}O_{2}^{-}+H_{2}O-->HC_{2}H_{3}O_{2}+OH^{-},
kb=kw/ka=5.56*10^{-10}**

**inital 0.25 0 0**

**equlibrium 0.25-x x x**

**thus,**

**let x<<0.25**

**thus**

**x=1.18*10 ^{-5} M**

**[OH ^{-}]=x=1.18*10^{-5} M**

**pOH=-log([OH-])=4.92**

**pH+pOH=14**

**thus,pH=14-4.92=9.08**

#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).

Consider a solution of 2.0 M HCN and 1.0 M NaCN (Ka for HCN =
6.2 x 10-10). Which of the following statements is true?
a) The solution is not a buffer because [HCN] is not equal to
[CN-]
b) The pH will be below 7.00 because the concentration of the
acid is greater than that of the
base.
c) [OH-] > [H+]
d) The buffer will be more resistant to pH changes from addition
of strong acid than to...

Find the pH of a 0.250 M solution of NaC2H3O2 . (The
Ka value of HC2H3O2 is 1.80×10−5).
Express your answer numerically to four significant figures.
Please Help. Thank you!!!

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

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

The acid-dissociation constant of hydrocyanic acid (HCN) at
25.0°C is 4.9 × 10-10. What is the pH of an aqueous
solution of 0.030 M sodium cyanide (NaCN)?

Equal volumes of 0.1 M NH3 (Kb= 1.8x10^-5) and 0.1 M HCN (Ka=
4.9x10^-10) are mixed together. will the resulting solution be
acidic, basic, or neutral? PLEASE SHOW WORK

a)You obtain a 0.817 M solution of NH4Cl. Knowing that the Kb of
NH3 is 1.8 * 10-5, what is the Ka of NH4+?
b) What is the [H+] in
the solution in part a?
c) What is the pH of the solution in part
a?

1. The Ka for HF is 6.9 x 10-4.
What is the value of Kb for F-? (Show your calculations)
2. The Kb for NH3 is 1.8 x 10-5. What is the value of Ka for
NH4+? (Show your calculations)
3. From what acid and what base were the following salts made
and what would you expect the pH of the salt solution to be (>
7, =7 or <7)? Why?
Rb3PO4
NH4I

Find the pH of a 0.342 M NaC2H3O2 solution. (The Ka of acetic
acid, HC2H3O2, is 1.8×10−5.)

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