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ella [17]
3 years ago
15

the student performed a second trial and accidentally added more sodium hydroxide to the flask than was needed to reach the end

point but recorded the final volume. would this error increase, decrease or have no effect on the calculated acid concentration for the second trial?
Chemistry
1 answer:
Harrizon [31]3 years ago
4 0

A greater volume of acid will be needed to neutralize the sodium hydroxide thereby making the calculated concentration of acid for the second trial to decrease.

Titration is usually a way of determining the concentration of an unknown solution by accurately measuring the volume of a known solution that reacts completely with the unknown solution.

The point when reaction is completed is indicated by the end point of the reaction. If more base (sodium hydroxide) is added to the flask, then more volume of acid will be needed to neutralize it thereby making the calculated concentration of acid for the second trial to decrease.

Learn more: brainly.com/question/1527403

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<u>Answer:</u> The concentration of original sulfuric acid solution is 1.62 M

<u>Explanation:</u>

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

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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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