Question

Determine the pH of each of the following solutions. 7.00×10−2 M HNO3, 0.160 M HNO2, 2.10×10−2 M KOH, 0.245 M CH3NH3I, 0.318 M KC6H5O.

I am not getting the correst answer to these.... please help and show work. Thank you

Answer 1

Determine the pH of each of the following solutions. 7.00×10−2 M HNO3, 0.160 M HNO2, 2.10×10−2 M KOH, 0.245 M CH3NH3I, 0.318 M KC6H5O.

Solution :-

1). 7.00×10−2 M HNO3

HNO3 is the strong acid therefore it dissociate 100 %

So the H+ concentration = 7.00*10^-2 M

pH= -log [H+]

    = - log [7.00*10^-2 ]

   = 1.15

2) 0.160 M HNO2

HNO2 is weak acid so need to use the Ka of the HNO2 to calculate the H+

HNO2 + H2O   -------- > H3O+   + NO2-

0.160                                 0             0

-x                                     +x             +x

0.160 –x                         x                  x

Ka= [H3O+][NO2-]/[HNO2]

4.5*10^-4 = [x][x]/[0.160-x]

4.5*10^-4 * 0.160-x = x^2

Solving for x we get

0.00826 = x = [H3O+]

pH= -log [H3O+]

pH= -log [0.00826]

pH= 2.08

3) 2.10×10−2 M KOH

KOH is strong base therefore it dissociate 100 %

So the [OH-] = 2.10*10^-2 M

pOH= - log [2.10*10^-2]

pOH= 1.68

pOH + pH= 14

pH= 14 – pOH

    = 14 – 1.68

   = 12.32

4) 0.245 M CH3NH3I

Need to use the ka of the methylamine

Kb of methyl amine = 4.38*10^-4

Ka= Kw/ kB

    = 1*10^-14 / 4.38*10^-4

= 2.28 *10^-11

CH3NH3+   + H2O   ------ > CH3NH2 + H3O+

0.245                                       0                 0

-x                                            +x                 +x

0.245-x                                x                       x

Ka= [CH3NH2][H3O+] /[CH3NH3+]

2.28*10^-11 = [x][x]/[0.245-x]

Since kb is very small then we can neglect the x from the denominator

2.28*10^-11 = [x][x]/[0.245]

2.28*10^-11 * 0.245 = x^2

5.586*10^-12 = x^2

Taking square root of both sides we get

2.36*10^-6 = x =[H3O+]

pH= -log [H3O+]

pH= -log [2.36*10^-6]

pH= 5.63

5) 0.318 M KC6H5O

C6H5O- + H2O ------ > C6H6O + OH-

0.318                                 0              0

-x                                       +x           +x

0.318-x                                x           x

Kb of phenolate ion = 7.69*10^-5

Kb = [C6H6O][OH-] /[C6H5O-]

7.69*10^-5 = [x][x]/[0.318-x]

7.69*10^-5 * 0.318 = x^2

2.44*10^-5 = x^2

Taking square root of both sides we get

0.00494 = x =[OH-]

pOH = -log [OH-]

pOH= -log [ 0.00494]

pOH= 2.31

pH= 14 – pOH

    = 14-2.31

   = 11.69