Question

Arrange the following 0.10 M
solutions from lowest to highest pH:
NaF, NaC2H3O2, C5H5NHCl, KOH, HCN.
(Ka for HCN is 6.2 × 10–10; Ka for HF is 7.2 × 10–4; Ka for HC2H3O2 is 1.8 × 10–5; and Kb for C5H5N is 1.7 × 10–9)
A) C5H5NHCl, HCN, NaF, NaC2H3O2,KOH
B) NaF, NaC2H3O2, HCN, C5H5NHCl, KOH
C) KOH, NaC2H3O2, NaF, HCN, C5H5NHCl
D) HCN, C5H5NHCl, NaF, NaC2H3O2, KOH
E) none of these

Answer 1

Let's calculate the pH of each of these:

NaF---> Na+ + F-

            F- + H2O <----> HF+OH-

I         0.10                    0      0

C      -x                         +x     +X

E     0.10-x                   x           x

x²/0.10-x = kb

kbXka = kw

ka = 10-¹⁴ / 7.2 X10^-4 = 1.38 X 10 ^-11

x²/0.10-x = 1.38 X 10 ^-11

X = [OH-] = 1.17 X 10^-6

p(oH) =-log(1.17 X 10^-6) = 5.93

pH = 14-5.93 = 8.07

B) 0.10 Na C2H3O2

C2H3O2- <------> HC2H3O2 +OH-

I         0.10                    0      0

C      -x                         +x     +X

E     0.10-x                   x           x

x²/0.10-x = kb

kbXka = kw

ka = 10-¹⁴ / 1.8X 10^-5 =0.55 X 10^-9

x²/0.10-x =0.55 X 10^-9

X = [OH-] = 7.4X 10^-6

p(oH) =-log(7.4 X 10^-6) = 5.130

pH = 14-5.130 = 8.87

C) C5H5NHCl

C5H5NH+ +H2O ----> C5H5N +H3O+

I         0.10                    0      0

C      -x                         +x     +X

E     0.10-x                   x           x

x²/0.10-x = ka

kbXka = kw

Ka =0.588 X 10^-5

X = [H3O+] =7.6 X 10^-4

pH = -log [H3O+] =3.119

D) 0.10 M KOH

[H+][OH-] = 10^-14

[H+] = 10^-13

pH = 13

E) 0.10 M HCN

HCN <----> H+ + CN-

I         0.10                    0      0

C      -x                         +x     +X

E     0.10-x                   x           x

x²/0.10-x = ka= 6.2X10^-10

X = 7.8x10^-6

pH = -log 7.8x10^-6 = 5.10

C5H5NHCl, HCN, NaF, NaC2H3O2, KOH is the correct order