Answer to Question #198095 in General Chemistry for Master j

Question

1. What is the pH of a 0.34 M solution of (weak acid) HF if the Ka = 6.8x10-4

?

2. If the pH of a weak acid solution is 9.5 and the original concentration of base was 0.30 M

A. what is the pOH?

B. what is the concentration of OH-

?

C. what is the equilibrium concentration of the base?

D. what is the Kb of the base?

3. Find the pH of a 0.325 M (weak acid) acetic acid solution. Ka = 1.8 x 10-5

.

4. Find the pH of a 0.056 M (weak acid)propionic acid solution (Ka = 1.4 x 10-5

).

5. Find the pH of a 0.075 M solution of (weak acid) formic acid. The acid dissociation

constant (Ka) for formic acid is 1.8 x 10-4

.

6. Find the pH of a 0.15 M solution of (weak base) ammonia, NH3. Kb = 1.8 X 10-5

7. Find the pH of a 0.600 M solution of (weak base) methylamine CH3NH2. Kb = 4.4 x 10–4

.

8. If the pH of (weak acid) HC3H5O2 is 4.2 and the Ka = 1.34x10-5

A. what is the equilibrium concentration of HC3H5O2?

B. what was the initial concentration of HC3H5O before dissociation?

Answer 1

1. K_a=[H⁺]×[F^-]/[HF]

6.8*10^-4=x*x/0.34-x from this eqution :

x²+6.8*10^-4 - 2.312*10^-4

x=0.0155 M ,

pH= -log [H⁺]=-log (0.0155)=1.8

2. BOH=B⁺+OH^-

pOH=14-9.5=4.5

[OH^-]=10^-pOH=10^-4.5=3.162*10^-5M

equlibrium concentration(BOH)=0.3--3.162*10^-5 =0.29996838M

K_b=[B⁺]×[OH^-]/[BOH]=3.162*10^-5 ×3.162*10^-5/0.29996838=1*10^-9

3.HA<-> H⁺+A^-

0.325 0 0

-x +x +x

0.325-x x x

K_a=[H⁺]*[A^-]/[HA]

1.8*10^-5=x*x/0.325-x

So x=0.00242M[H⁺] ,pH=-log [H⁺]=

-log(0.00242)=2.616

4.HA<-> H⁺+A^-

0.056 0 0

0.056-x +x +x

1.4*10^-5=x*x/0.056

So x=0.0008854M

pH=-log[0.0008854]=3.05

5.

1.8*10^-4=x*x/0.075

So x=0.003674M

pH=-log [0.003674]=2.4348

6.

1.8*10^-5=x*x/0.15

So x=0.001643M

pH=-log [0.001643]=2.7843

7.

4.4*10^-4=x*x/0.6

So x=0.016248M

pH=-log[0.016248]=1.7891

8.

pH=4.2

[H⁺]=10^-pH=10^-4.2=6.3*10^-5M

HA <--> H⁺ + A^-

6.3*10^-5 6.3*10^-5 6.3*10^-5

x-6.3*10^-5 +x +x

1.34*10^-5=6.3*10^-5*6.3*10^-5/x-6.3*10^-5

So x=0.0066M (HA)