Question

Suppose you are titrating a sulfuric acid solution of unknown concentration with a sodium hydroxide solution according to the equation H 2 S O 4 + 2 N a O H ⟶ 2 H 2 O + N a 2 S O 4 If you require 28.07 mL of 0.697 M NaOH solution to titrate 220.1 mL of H 2 SO 4 solution, what is the concentration of the H 2 SO 4 solution? Type answer:

Answer 1

Answer:

$M_{acid}=0.0444M$

Explanation:

Hello,

In this case, for the given reaction:

$H _2 S O _4 + 2 N a O H \rightarrow 2 _H 2 O + N a _2 S O_ 4$

We find a 1:2 molar ratio between the acid and the base respectively, for that reason, at the equivalence point we find:

$2*n_{acid}=n_{base}$

That in terms of concentrations and volumes we can compute the concentration of the acid solution:

$2*M_{acid}V_{acid}=M_{base}V_{base}\\\\M_{acid}=\frac{M_{base}V_{base}}{2*V_{acid}}=\frac{0.697M*28.07mL}{2*220.1mL}\\ \\M_{acid}=0.0444M$

Best regards.

Answer 2

The concentration of the sulfuric acid solution (H₂SO₄) is 0.0444 M

From the question,

We are to determine the concentration of the H₂SO₄ solution

Using the titration 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

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

The given balanced chemical equation for the reaction is

H₂SO₄ + 2NaOH ⟶ 2H₂O + Na₂SO₄

∴ $n_{A} = 1$

and $n_{B} = 2$

From the given information

$V_{A} = 220.1 \ mL$

$C_{B} = 0.697 \ M$

$V_{B} = 28.07 \ mL$

Putting the parameters into the equation, we get

$\frac{C_{A} \times 220.1}{0.697 \times 28.07}= \frac{1}{2}$

∴ $C_{A} = \frac{1 \times 0.697 \times28.07}{2\times 220.1}$

$C_{A} = \frac{19.56479}{440.2}$

$C_{A} = 0.0444 \ M$

Hence, the concentration of the sulfuric acid solution (H₂SO₄) is 0.0444 M

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