Chemistry, 23.06.2020 17:01 adriancastaneda

50cm3 of sodium hydroxide solution was titrated against a solution of sulfuric acid. The concentration of the sodium hydroxide solution was 20g/dm3. Work out the concentration of the acid in grams per litre if it took 25cm3 of acid to completely neutralise the alkali. The relative molecular mass of sulfuric acid is 98.

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Answer from: kingemzy

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 H₂SO₄

From the formula

$\frac{C_{A}V_{A} }{C_{B}V_{B} }=\frac{n_{A} }{n_{B} }$

Where

$C_{A}$ is the concentration of acid

$C_{B}$ is the concentration of base

$V_{A}$ is the volume of acid

$V_{B}$ is the volume of base

$n_{A}$ is the mole ratio of acid

$n_{B}$ is the mole ratio of base

From the question

$V_{A}= 25 \ cm^{3}$

$C_{B} = 20 \ g/dm^{3}$

Convert this to mol/dm³

Molar mass of NaOH = 40 g/mol

Using the formula

$Concentration\ in \ mol/dm^{3} =\frac{Concentration\ in \ g/dm^{3}}{Molar\ mass}$

∴ Concentration of the NaOH in mol/dm³ = $\frac{20}{40}$

Concentration of the NaOH in mol/dm³ = 0.5 mol/dm³

∴ $C_{B} = 0.5 \ mol/dm^{3}$

and

$V_{B} = 50 \ cm^{3}$

From the balanced chemical equation

$n_{A} = 1$

$n_{B} = 2$

Putting the parameters into the formula, we get

$\frac{C_{A} \times 25}{0.5 \times 50}= \frac{1}{2}$

Now, this becomes

$2 \times 25 \times C_{A}=0.5 \times 50 \times 1$

Then, we get

$50 C_{A} = 25$

∴ $C_{A} = \frac{25}{50}$

$C_{A} = 0.5 \ mol/dm^{3} = 0.5 \ mol/L$

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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Answer from: GachaSkylarUwU

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

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