Consider the following data on some weak acids and weak bases: Benoize Louisa, x

Question

Consider the following data on some weak acids and weak bases:

Benoize Louisa, x Secure | https://www-awh.aleks.com/al aleksogi/x/Islexe/1o u-lgNslkr7j8P3)H-IBx4uFZeigla Predicting the qualitative acid-base properties of salts Consider the following data on some weak acids and weak bases: acidKa formula base name formula hydrofluoric acid HF 6810 ammonia NH 1.8 x 10 hydrocyanic acid HCN 4.9 10 pyridine CSH N 1.7x 10 -9 Use this data to rank the following solutions in order of increasing pH. In other words the solution that will have the next lowest pH, and so on. solution pH 0.1 M NaCN 0.1 M KNO 0.3 M NH.Br choose one choose one choose one choose one 0.1 M CSHsNHC Explanation Check Type here to search

Explanation / Answer

for NaCN:

use:

Kb = Kw/Ka

Kw is dissociation constant of water whose value is 1.0*10^-14 at 25 oC

Kb = (1.0*10^-14)/Ka

Kb = (1.0*10^-14)/4.9*10^-10

Kb = 2.041*10^-5

CN- dissociates as

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

0.1 0 0

0.1-x x x

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

Kb = x*x/(c-x)

Assuming x can be ignored as compared to c

So, above expression becomes

Kb = x*x/(c)

so, x = sqrt (Kb*c)

x = sqrt ((2.041*10^-5)*0.1) = 1.429*10^-3

since x is comparable c, our assumption is not correct

we need to solve this using Quadratic equation

Kb = x*x/(c-x)

2.041*10^-5 = x^2/(0.1-x)

2.041*10^-6 - 2.041*10^-5 *x = x^2

x^2 + 2.041*10^-5 *x-2.041*10^-6 = 0

This is quadratic equation (ax^2+bx+c=0)

a = 1

b = 2.041*10^-5

c = -2.041*10^-6

Roots can be found by

x = {-b + sqrt(b^2-4*a*c)}/2a

x = {-b - sqrt(b^2-4*a*c)}/2a

b^2-4*a*c = 8.164*10^-6

roots are :

x = 1.418*10^-3 and x = -1.439*10^-3

since x can't be negative, the possible value of x is

x = 1.418*10^-3

use:

pOH = -log [OH-]

= -log (1.418*10^-3)

= 2.8482

use:

PH = 14 - pOH

= 14 - 2.8482

= 11.15

for KNO3:

This is salt of strong acid HNO3 and strong base KOH

So, pH will be 7.0

for NH4Br:

use:

Ka = Kw/Kb

Kw is dissociation constant of water whose value is 1.0*10^-14 at 25 oC

Ka = (1.0*10^-14)/Kb

Ka = (1.0*10^-14)/1.8*10^-5

Ka = 5.556*10^-10

NH4+ + H2O -----> NH3 + H+

0.1 0 0

0.1-x x x

Ka = [H+][NH3]/[NH4+]

Ka = x*x/(c-x)

Assuming x can be ignored as compared to c

So, above expression becomes

Ka = x*x/(c)

so, x = sqrt (Ka*c)

x = sqrt ((5.556*10^-10)*0.1) = 7.454*10^-6

since c is much greater than x, our assumption is correct

so, x = 7.454*10^-6 M

So, [H+] = x = 7.454*10^-6 M

use:

pH = -log [H+]

= -log (7.454*10^-6)

= 5.123

for C5H5NHCl:

use:

Ka = Kw/Kb

Kw is dissociation constant of water whose value is 1.0*10^-14 at 25 oC

Ka = (1.0*10^-14)/Kb

Ka = (1.0*10^-14)/1.7*10^-9

Ka = 5.882*10^-6

C5H5NH+ + H2O -----> C5H5N + H+

0.1 0 0

0.1-x x x

Ka = [H+][C5H5N]/[C5H5NH+]

Ka = x*x/(c-x)

Assuming x can be ignored as compared to c

So, above expression becomes

Ka = x*x/(c)

so, x = sqrt (Ka*c)

x = sqrt ((5.882*10^-6)*0.1) = 7.67*10^-4

since c is much greater than x, our assumption is correct

so, x = 7.67*10^-4 M

So, [H+] = x = 7.67*10^-4 M

use:

pH = -log [H+]

= -log (7.67*10^-4)

= 3.12

Answer:

0.1 M NaCN: 4

01 M KNO3: 3

0.1 M NH4Br: 2

0.1 M C5H5NHCl: 1

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