5) Calculate the pH of each of the following solutions: (a) 0.0010 MHC (b) 0.76

Question

5) Calculate the pH of each of the following solutions: (a) 0.0010 MHC (b) 0.76 MKOH: (c) 0.0187 M CHCOOH, (K. for CHC00H-1.8 × 10-5); (d) 0.040 M H2SO. (Hint: H2SO. is a strong acid%, for HS04-13x 10-2.) (6) Calculate the percent ionization of hydrofluoric acid at the following concentrations: (a) 0.60 M (b) 0.0046 M (c 0.00028 M (d) As the solution becomes more dilute, what happens to the percent ionization? (K, for hydro fluoric acid = 7. 1 × 10 4, percent ionization-ionized acid concentration at equilibrium 100% initial concentration ofacid

Explanation / Answer

Q5.

a)

[H+] = [HCl] = 0.001 M

then

pH = -log(H+]) = -log(0.001)  

pH = 3

b)

KOH --> [OH-] = [KOH] = 0.76

pOH = -log(OH) = -log(0.76) = 0.1191

pH = 14-pOH = 14-0.1191

pH = 13.8809

c)

First, assume the acid:

HF

to be HA, for simplicity, so it will ionize as follows:

HA <-> H+ + A-

where, H+ is the proton and A- the conjugate base, HA is molecular acid

Ka = [H+][A-]/[HA]; by definition

initially

[H+] = 0

[A-] = 0

[HA] = M;

the change

initially

[H+] = + x

[A-] = + x

[HA] = - x

in equilbrirum

[H+] = 0 + x

[A-] = 0 + x

[HA] = M - x

substitute in Ka

Ka = [H+][A-]/[HA]

Ka = x*x/(M-x)

x^2 + Kax - M*Ka = 0

if M = 0.0187 M; then

x^2 + (1.8*10^-5)x - 0.0187 *(1.8*10^-5) = 0

solve for x

x =0.01434

substitute

[H+] = 0 + 5.712*10^-4 = 0 + 5.712*10^-4 M

pH = -log(H+) = -log(0 + 5.712*10^-4) = 3.2432

d)

[H+]1st proton = 0.04 M

after this

[H+] = 0.04 + x

[SO4-2] = x

[HSO4-] = 0.04-x

Ka = [H+][SO4-2]/[HSO4-]

(1.3*10^-2) = x*(0.04+x)/(0.04-x)

x^2 + 0.04x = 0.013*0.04 - 0.013x

x^2 + (0.04+0.013)x - (0.013*0.04) = 0

x = 0.00846

[H+] = 0.04 + 0.00846 = 0.04846

pH = -log(0.04846) = 1.31

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