Calculate the volume of 0.5 mol dm\(^{-3}\) H\(_2\)SO\(_4\) that is neutralized by 25 cm\(^3\) of 0.1 mol dm\(^3\) NaOH?

godstamtam

7 years ago

CA VA /CB VB = na/nb
H2SO4+ 2NaOH—— Na2SO4+2H2O
1 : 2
CA=0.5
VA=?
CB=0.1
VB=25
na=1
nb=2
...0.5×VA/0.1×25= 1/2
0.1×25×1=0.5×2×CB
...CB=0.1×25×1/0.5×2
CB=2.5

flinus

4 years ago

The equation is ;
H2SO4 + 2NaOH •••••••••> NaSO4 + 2H2O
| |
1mol 2mol

Using the formula: CaVa/CbVb = Na/Nb

Ca = 0.5mol|dm³
VA = ?
Cb = 0.1mol|dm³
Vb = 25cm³
Na = 1
Nb = 2

Using the formula
CaVa / CbVb = Na / Nb
Therefore
VA = CbVbNa / CaNb

= 0.1* 25 * 1 / 0.5 * 2
= 2.5 / 1
= 2.5
Hence VA= 2.5cm³

tyjesma

11 years ago

CaVa/CbVb=na/nb....Ca=0.5, Va=?, Cb=0.1, Vb=25cm....



NaOH + H2SO4 = Na2So4 + H2O,



therefore ratio of no of moles of Acid to Base is 1:1...



CaVa x nb=CbVb x na...Va=CbVb x na/Ca x na ...



Va=0.1 x 25 x 1/0.5 x 1= 5

vickyf

9 years ago

no ur equation is nt balanced....it is 2:1...xo d ans..is 2.5...myschool God bless u

frankjoe033

3 years ago

H2SO4 + 2NAOH== NA2SO4+2H20
1Mol 2mol 1mol 2mol
CaVa/CbVb=1/2
0.5Va=1
0.1×25=2
2×0.5Va=1×0.1×25
1va=2.5
Va=2.5

Grachiev

1 year ago

nawa oo

Ayomi424

3 years ago

2NAOH+H2So4 > 2NASO4 +2H2
2. : 1
NaOH=Base
H2So4=Acid

Concentration of acid =0.5mol dm-3
concentration of base =0.1mol dm-3
volume of acid = ?
volume of base =25Cm2

Ca Va/Cb Vb =Na/Nb
0.5*Va/0.1*25 =1/2
cross multiply
Va=0.1*25/0.5*2
Va=2.5/1
Va=2.5cm3

Ebighebisolomon

1 year ago

i need explanation to this

Ebighebisolomon

1 year ago

what did CA stands for

RiverHeights

2 years ago

The explanation video was very good

Komikwhyke

2 years ago

To calculate the volume of
0.5
 
mol dm
−
3
0.5mol dm
−3
H₂SO₄ neutralized by
25
 
cm
3
25cm
3
of
0.1
 
mol dm
−
3
0.1mol dm
−3
NaOH, we can use the concept of stoichiometry.

The balanced chemical equation for the neutralization reaction between sulfuric acid (H₂SO₄) and sodium hydroxide (NaOH) is:

H
2
SO
4
+
2
NaOH
→
Na
2
SO
4
+
2
H
2
O
H
2
​
SO
4
​
+2NaOH→Na
2
​
SO
4
​
+2H
2
​
O

From the equation, we can see that 1 mole of H₂SO₄ reacts with 2 moles of NaOH.

Given that the concentration of NaOH is
0.1
 
mol dm
−
3
0.1mol dm
−3
and the volume used is
25
 
cm
3
25cm
3
(which can be converted to
0.025
 
dm
3
0.025dm
3
), we can calculate the number of moles of NaOH used:

moles of NaOH
=
concentration
×
volume
moles of NaOH=concentration×volume
=
0.1
 
mol dm
−
3
×
0.025
 
dm
3
=0.1mol dm
−3
×0.025dm
3

=
0.0025
 
moles
=0.0025moles

According to the stoichiometry of the reaction, 1 mole of H₂SO₄ reacts with 2 moles of NaOH. Therefore, the number of moles of H₂SO₄ neutralized is half the number of moles of NaOH used.

moles of H
2
SO
4
=
1
2
×
moles of NaOH
moles of H
2
​
SO
4
​
=
2
1
​
×moles of NaOH
=
1
2
×
0.0025
 
moles
=
2
1
​
×0.0025moles
=
0.00125
 
moles
=0.00125moles

Now, we can use the concentration of H₂SO₄ to calculate the volume of H₂SO₄ required to neutralize this amount of moles:

volume
=
moles
concentration
volume=
concentration
moles
​

=
0.00125
 
moles
0.5
 
mol dm
−
3
=
0.5mol dm
−3

0.00125moles
​

=
0.0025
 
dm
3
=0.0025dm
3


Finally, we can convert this volume from dm³ to cm³:

volume
=
0.0025
 
dm
3
×
1000
 
cm
3
/
dm
3
volume=0.0025dm
3
×1000cm
3
/dm
3

=
2.5
 
cm
3
=2.5cm
3


Therefore, 2.5 cm³ of
0.5
 
mol dm
−
3
0.5mol dm
−3
H₂SO₄ is neutralized by
25
 
cm
3
25cm
3
of
0.1
 
mol dm
−
3
0.1mol dm
−3
NaOH.

Eaglelife

1 year ago

SHOW THE EXPLANATION