It takes 40 mL of 0.10 M of hydrochloric acid (HCl) to neutralize 50 mL of sodium hydroxide (NaOH) for the following chemical reaction

HCl + NaOH → NaCl + H2O

The concentration of sodium hydroxide is _____.
0.125 M
0.08 M
0.10 M
0.04 M

Answers

Answer 1

Answer:

0.08 M

Explanation:

here...

1. water and a salt

2. Magnesium hydroxide reacts with HCl to produce a solution that is neutral.

3. titrant

4. 0.08 M

5. indicator

{quizlet: captncrun}

Answer 2

The concentration of sodium hydroxide is 0.08 M. The correct option is the second option 0.08 M

From the question,

We are to determine the concentration of sodium hydroxide

The given reaction is

HCl + NaOH → NaCl + H₂O

This means

1 mole of HCl is required to completely neutralize 1 mole of NaOH

To determine the concentration of sodium hydroxide,

We will use 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 give information

[tex]C_{A} = 0.10 \ M[/tex]

[tex]V_{A} = 40 \ mL[/tex]

[tex]V_{B} = 50 \ mL[/tex]

From the balanced chemical equation

[tex]n_{A} = 1[/tex]

[tex]n_{B} = 1[/tex]

Putting the above parameters into the formula, we get

[tex]\frac{0.10 \times 40}{C_{B} \times 50} = \frac{1}{1}[/tex]

Then

[tex]C_{B} \times 50 = 0.10 \times 40[/tex]

∴ [tex]C_{B}=\frac{0.1 \times 40}{50}[/tex]

[tex]C_{B}=\frac{4}{50}[/tex]

[tex]C_{B}=0.08[/tex]

Hence, the concentration of sodium hydroxide is 0.08 M. The correct option is the second option 0.08 M

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