Respuesta :
Answer:
49 g/L is the concentration of the acid
Explanation:
Firstly, we proceed to write the equation of reaction.
2NaOH + H2SO4 ——-> Na2SO4 + 2H2O
We can see that 1 mole of the base reacted with two moles of the acid.
kindly note that dm^3 is same as liter
Firstly, we need to get the concentration of the reacted sulphuric acid in g/L
we use the simple titration equation below;
CaVa/CbVb = Na/Nb
From the question;
Ca = ?
Va = 25 cm^3
Cb = 20 g/L
we convert this to concentration in mol/L
Mathematically, that is concentration in g/L divided by molar mass in g/mole
molar mass of NaOH = 40 g/mol
so we have; 20g/L / 40 = 0.5 mol/L
Vb = 50 cm^3
Na = 1
Nb = 2
Where C represents concentrations, V volumes and N , number of moles
Now, substitute the values;
Ca * 25/0.5 * 50 = 1/2
25Ca/25 = 0.5
So Ca = 0.5 mol/L
Now to get the concentration of H2SO4 in g/L
What we do is to multiply the concentration in mol/L by molar mass in g/mol
That would be 0.5 * 98 = 49 g/L
The concentration of the acid in grams per litre is 49g/L
From the question,
We are to determine concentration of the sodium hydroxide solution.
First, we will write a balanced chemical equation for the reaction.
The balanced chemical equation for the reaction is
2NaOH + H₂SO₄ → Na₂SO₄ + 2H₂O
This means,
2 moles of NaOH is required to completely neutralize 1 mole of Hâ‚‚SOâ‚„
From the formula
[tex]\frac{C_{A}V_{A} }{C_{B}V_{B} }=\frac{n_{A} }{n_{B} }[/tex]
Where
[tex]C_{A}[/tex] is the concentration of acid
[tex]C_{B}[/tex] is the concentration of base
[tex]V_{A}[/tex] is the volume of acid
[tex]V_{B}[/tex] is the volume of base
[tex]n_{A}[/tex] is the mole ratio of acid
[tex]n_{B}[/tex] is the mole ratio of base
From the question
[tex]V_{A}= 25 \ cm^{3}[/tex]
[tex]C_{B} = 20 \ g/dm^{3}[/tex]
Convert this to mol/dm³
Molar mass of NaOH = 40 g/mol
Using the formula
[tex]Concentration\ in \ mol/dm^{3} =\frac{Concentration\ in \ g/dm^{3}}{Molar\ mass}[/tex]
∴ Concentration of the NaOH in mol/dm³ = [tex]\frac{20}{40}[/tex]
Concentration of the NaOH in mol/dm³ = 0.5 mol/dm³
∴ [tex]C_{B} = 0.5 \ mol/dm^{3}[/tex]
and
[tex]V_{B} = 50 \ cm^{3}[/tex]
From the balanced chemical equation
[tex]n_{A} = 1[/tex]
[tex]n_{B} = 2[/tex]
Putting the parameters into the formula, we get
[tex]\frac{C_{A} \times 25}{0.5 \times 50}= \frac{1}{2}[/tex]
Now, this becomes
[tex]2 \times 25 \times C_{A}=0.5 \times 50 \times 1[/tex]
Then, we get
[tex]50 C_{A} = 25[/tex]
∴ [tex]C_{A} = \frac{25}{50}[/tex]
[tex]C_{A} = 0.5 \ mol/dm^{3} = 0.5 \ mol/L[/tex]
Now, we will determine the concentration in g/L
From the formula,
Concentration in g/L = Concentration in mol/L × Molar mass
From the question,
Molar mass of sulfuric acid = 98
∴ Concentration of the acid in g/L = 0.5 × 98
Concentration of the acid in g/L = 49 g/L
Hence, the concentration of the acid in grams per litre is 49g/L
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