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3 years ago

WHAT ELEMENT IS THIS

Chemistry

1 answer:

LenaWriter [7]3 years ago

4 0

Answer: emission spectrum

Explanation:

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A chemist needs to determine the concentration of a sulfuric acid solution by titration with a standard sodium hydroxide solutio

Zigmanuir [339]

Answer: The concentration of original sulfuric acid solution is 1.62 M

Explanation:

Let the original concentration of sulfuric acid be 'x' M

To calculate the molarity of the diluted solution, we use the equation:

$M_1V_1=M_2V_2$

where,

$M_1\text{ and }V_1$ are the molarity and volume of the concentrated sulfuric acid solution

$M_2\text{ and }V_2$ are the molarity and volume of diluted sulfuric acid solution

We are given:

$M_1=xM\\V_1=25.00mL\\M_2=?M\\V_2=250.0mL$

Putting values in above equation, we get:

$x\times 25.00=M_2\times 250.0\\\\M_2=\frac{x\times 25.0}{250}=\frac{x}{10}$

Now, to calculate the concentration of acid, we use the equation given by neutralization reaction:

$n_1M_1V_1=n_2M_2V_2$

where,

$n_1,M_1\text{ and }V_1$ are the n-factor, molarity and volume of acid which is $H_2SO_4$

$n_2,M_2\text{ and }V_2$ are the n-factor, molarity and volume of base which is NaOH.

We are given:

$n_1=2\\M_1=\frac{x}{10}M\\V_1=10.00mL\\n_2=1\\M_2=0.1790M\\V_2=18.07mL$

Putting values in above equation, we get:

$2\times \frac{x}{10}\times 10.00=1\times 0.1790\times 18.07\\\\x=\frac{1\times 0.1790\times 18.07\times 10}{2\times 10.00}=1.62M$

Hence, the concentration of original sulfuric acid solution is 1.62 M

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3 years ago