Question

Determine the pH of a 0.200 M solution of hydrofluoric acid (HF) (Ka = 6.3 x 10^-4)

a. 4.20

b. 3.90

c. 2.02

d. 0.69

*please show process*

Answer #1

HF dissociates as:

HF -----> H+ + F-

0.2 0 0

0.2-x x x

Ka = [H+][F-]/[HF]

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 ((6.3*10^-4)*0.2) = 1.122*10^-2

since x is comparable c, our assumption is not correct

we need to solve this using Quadratic equation

Ka = x*x/(c-x)

6.3*10^-4 = x^2/(0.2-x)

1.26*10^-4 - 6.3*10^-4 *x = x^2

x^2 + 6.3*10^-4 *x-1.26*10^-4 = 0

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

a = 1

b = 6.3*10^-4

c = -1.26*10^-4

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 = 5.044*10^-4

roots are :

x = 1.091*10^-2 and x = -1.154*10^-2

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

x = 1.091*10^-2

So, [H+] = x = 1.091*10^-2 M

use:

pH = -log [H+]

= -log (1.091*10^-2)

= 1.96

Answer: d

a. Calculate the pH of a 0.538 M aqueous
solution of hydrofluoric acid
(HF, Ka =
7.2×10-4).
b. Calculate the pH of a 0.0242 M aqueous
solution of nitrous acid
(HNO2, Ka =
4.5×10-4).

A.) Find the pH of a 0.338 M NaF solution. (The Ka of
hydrofluoric acid, HF, is 3.5×10?4.)
B.) Determine the [OH?] of a 0.30 M solution of NaHCO3. and
Determine the pH of this solution.
C.)Find the [OH?] of a 0.46 M methylamine
(CH3NH2) solution. (The value of Kb for methylamine
(CH3NH2) is 4.4×10?4.) include units
D.) Find the pH of a 0.46? M methylamine
(CH3NH2) solution.

Hydrofluoric acid, HF, is a weak acid with an acid-dissociation
constant (Ka) of 6.3 ✕ 10−4. If
0.023 L of 0.17 M HF is titrated with 0.16 M
NaOH, a strong base, determine the pH at the equivalence point.
(The ion-product constant of water isKw = 1.01
✕ 10−14 at 25°C.)

Determine the pH during the titration of 25.7
mL of 0.304 M hydrofluoric acid
(Ka = 7.2×10-4) by
0.478 M KOH at the following
points.
(a) Before the addition of any KOH _____
(b) After the addition of 4.20 mL of
KOH _____
(c) At the half-equivalence point (the titration midpoint)
_____
(d) At the equivalence point _____
(e) After the addition of 24.5 mL of
KOH _____

Find the pH of a 0.300 M HF (Ka=6.3×10−4) solution.
Find the percent dissociation of a 0.300 M HF solution

If a solution of HF (Ka = 6.8 x 10-4) has a pH of 3.65,
calculate the initial concentration of hydrofluoric acid.

A sample of 0.250 M hydrofluoric acid (HF) is reacted with 0.150
M sodium hydroxide (NaOH) solution. Ka = 6.8 x 10-4 for HF. What is
the pH when you mix 25.0 mL of HF with 50.0 mL of NaOH.

The acid disassociation constant, Ka, for hydrofluoric acid is
6.8 x 10−4. Given a 0.020 M solution of hydrofluoric acid, answer
the following questions. HF(aq) H+(aq) + F−(aq)
A. Write the equilibrium expression.
B. What is the concentration of acid, H+, in the equilibrium
solution?

Calculate the pH of a 0.150 M solution of benzoic acid
(C6H5CO2H) if
Ka= 6.3 x 10-5for the
acid. C6H5CO2H(aq) ⇌ H+(aq)
+ C6C5CO2-(aq)

Hydrofluoric acid, HF is a weak acis with Ka= 3.5X10^-4. It
dissociates according to
HF(aq) H^+(aq) + F^-(aq)
a. What is the pH of a buffer solution which is 0.30M in F- and
0.60M in HF?
b. What will be the pH of the resulting solution when 2.0mL of
0.5M HCl are added to 50mL of this (from part a) buffer
solution
c. What will be the pH of the resulting solution when 2.0mL of
0.5M NaOH are added...

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