Question #198095

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?

Expert's answer

1. K_{a}=[H^{+}]×[F^{-}]/[HF]

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

x^{2}+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^{-5}M

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^{-5}M

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)

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