Consider the following data on some weak acids and weak bases K a HCH3CO2 1.8 x1

Question

Consider the following data on some weak acids and weak bases

Ka

HCH3CO2

1.8 x105

HCN

4.9 x 1010

Kb

C5H5N

1.7 x 109

NH3

1.8 x 105

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.

0.1M NaCH3CO2

0.1M NH4Br

0.1M NaBr

0.1M KCN

acid

Ka

name formula acetic acid

HCH3CO2

1.8 x105

hydrocyanic acid

HCN

4.9 x 1010

base

Kb

name formula pyridine

C5H5N

1.7 x 109

ammonia

NH3

1.8 x 105

Explanation / Answer

1)

0.1M NaCH3CO2

NaCH3CO2 ------------------- Na+ + CH3CO2-

CH3CO2-   + H2O ----------------- CH3COOH + OH-

0.1                                              0                 0

- x                                              +x                 +x

Ka = 1.8x10^-5

KaxKb= Kw                Kw=ionic product of water = 1,0.x10^-14

Kb= Kw/Ka= 1.0x10^-14/1.8x10^-5 = 5.56x10^-10

Kb = [CH3COOH][OH-]/[CH3COO-]

5,56x10^-10 = x*x/(0.1-x)

5.56x10^-10 x0.1= x^2

x=0.746 x10^-5

[OH-]= 0.746x10^-5M

-log(OH-)= -log(0.746x10^-5)

POH= 5.12

PH+POH= 14

PH= 14 - 5.12=8.88

PH= 8.88

2)

0.1M NH4Br

NH4Br----------------------- NH4+ + Br-

NH4+ + H2O ------------ NH4OH + H+

0.1                                0           0

-x                                   +x        +x

0.1-x                              +x        +x

Kb= 1.8x10^-5

Ka= Kw/Kb

Ka 1.0x10^-14/1.8x10^-5

Ka= 5.56x10^-10

Ka = [NH4OH][H+]/[NH4+]

5.56x10^-10 = x*x/(0.1-x)

x^2= 5.56x10^-10x0.1

x=0.746x10^-5

[H+] = 0.746x10^-5

-log(H+) = -log(0.746x10^-5)

PH= 5.12

3)

0.1M NaBr

this is salt of both strong acid HBr and strong Base NaOH. so there is no hydrolysis takes place.Hence solution is neutral in nature. Hence PH =7

4)

0.1M KCN

KCN ---------------------- K+ + CN-

CN- H2O ------------ HCN + OH-

0.1                          0        0

-x                            +x       +x

0.1-x                        +x        +x

Ka= 4.9x10^-10

Kb= 2.04x10^-5

Kb= [HCN][OH-]/[CN-]

2.04x10^-5 = x*x/(0.1-x)

x = 1.43x10^-3

[OH-] = 1.43x10^-3

-log[OH-] = -log(1.43x10^-3)

POH= 2.84

PH= 14 -2.84

PH=11.16

0.1M KCN > 0.1M NaCH3COO > 0.1M NaBr > 0.1 M NH4Br

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