100ml of 0.1M CH,COOH is mixed with
50ml of 0.1M NaOH solution and pH of
the resulting solution is 5. The change in
pH if 100ml of 0.05M NaOH is added in
the above solution is
1) 1.30 2) 4.74 3) 5 4) 3.8
Which
the following ielowie Acid
Answers
Explanation:
A mixture containing 100 ml of 0.1 M acetic acid and 50 ml of 0.1 M NaOH is acidic buffer soluion. Its pH is given by the expression pH=pK
a
+log
acid
salt
.
Substitute values in the above expression.
5=pK
a
+log
0.1×0.05
0.1×0.05
or pK
a
=5.
When 100 ml of 0.05 M NaOH is added, the acid is completely neutralized and the solution contains salt sodium acetate.
The expression for the hydrogen ion concentration of the salt of weak acid and strong base is pH=
2
1
(pK
w
+pK
a
+logc).
Substitute values in the above expression.
pH=
2
1
(14+5+log(
0.25
0.1
))=8.8.
Hence, the change in pH is ΔpH=8.8−5=3.8.
Answer:
You are dealing with a neutralization reaction that takes place between acetic acid,
CH
3
COOH
, a weak acid, and sodium hydroxide,
NaOH
, a strong base.
Now, the pH of the resulting solution will depend on whether or not the neutralization is complete or not.
If the neutralization is not complete, more specifically if the acid is not completely neutralized, you will have a buffer solution that will contain acetic acid and its conjugate base, the acetate anion..
It's important to note that at complete neutralization, the pH of the solution will not be equal to
7
. Even if you neutralize the weak acid completely, the solution will be left with its conjugate base, which is why you can expect its pH to be greater than
7
.
So, the balanced chemical equation for this reaction is - I'll show you the ionic equation
CH
3
COOH
(aq]
+
OH
−
(aq]
→
CH
3
COO
−
(aq]
+
H
2
O
(l]
Notice that
1
mole of acetic acid will react with
1
mole of sodium hydroxide, shown here as hydroxide anions,
OH
−
, to produce
1
mole of acetate anions,
CH
3
COO
−
.
Use the molarities and volumes of the two solutions to determine how many moles of each you're adding
c
=
n
V
⇒
n
=
c
⋅
V
n
acetic
=
0.20 M
⋅
25.00
⋅
10
−
3
L
=
0.0050 moles CH
3
COOH