Question

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

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

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

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 give information

$C_{A} = 0.10 \ M$

$V_{A} = 40 \ mL$

$V_{B} = 50 \ mL$

From the balanced chemical equation

$n_{A} = 1$

$n_{B} = 1$

Putting the above parameters into the formula, we get

$\frac{0.10 \times 40}{C_{B} \times 50} = \frac{1}{1}$

Then

$C_{B} \times 50 = 0.10 \times 40$

∴ $C_{B}=\frac{0.1 \times 40}{50}$

$C_{B}=\frac{4}{50}$

$C_{B}=0.08$

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

Learn more here: brainly.com/question/14753195