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A chemist uses [tex]$0.25 \, L$[/tex] of [tex]$2.00 \, M \, H_2SO_4$[/tex] to completely neutralize [tex][tex]$2.00 \, L$[/tex][/tex] of a solution of [tex]NaOH[/tex]. The balanced chemical equation of the reaction is given below.

[tex]
2 \, NaOH + H_2SO_4 \rightarrow Na_2SO_4 + 2 \, H_2O
[/tex]

What is the concentration of [tex]NaOH[/tex] that is used?

[tex]
\begin{array}{l}
0.063 \, M \\
0.25 \, M \\
0.50 \, M \\
1.0 \, M
\end{array}
\]



Answer :

GinnyAnswer
Let's look at this problem step by step:

1. Identify the reactants and their volumes and molarities:
- Volume of [tex]\( H_2SO_4 \)[/tex] solution: 0.25 L
- Molarity of [tex]\( H_2SO_4 \)[/tex] solution: 2.00 M
- Volume of [tex]\( NaOH \)[/tex] solution: 2.00 L

2. Determine the number of moles of [tex]\( H_2SO_4 \)[/tex] used:
- Moles of [tex]\( H_2SO_4 \)[/tex] = Volume of [tex]\( H_2SO_4 \)[/tex] × Molarity of [tex]\( H_2SO_4 \)[/tex]
- Moles of [tex]\( H_2SO_4 \)[/tex] = 0.25 L × 2.00 M = 0.5 moles

3. Find the number of moles of [tex]\( NaOH \)[/tex] required for neutralization:
- According to the balanced chemical equation [tex]\( 2 NaOH + H_2SO_4 \rightarrow Na_2SO_4 + 2 H_2O \)[/tex], 1 mole of [tex]\( H_2SO_4 \)[/tex] reacts with 2 moles of [tex]\( NaOH \)[/tex].
- Therefore, moles of [tex]\( NaOH \)[/tex] required = 2 × moles of [tex]\( H_2SO_4 \)[/tex]
- Moles of [tex]\( NaOH \)[/tex] required = 2 × 0.5 moles = 1.0 mole

4. Calculate the molarity of the [tex]\( NaOH \)[/tex] solution:
- Molarity (M) = Moles of solute / Volume of solution in liters
- Molarity of [tex]\( NaOH \)[/tex] = Moles of [tex]\( NaOH \)[/tex] / Volume of [tex]\( NaOH \)[/tex]
- Molarity of [tex]\( NaOH \)[/tex] = 1.0 mole / 2.00 L = 0.5 M

Therefore, the concentration of [tex]\( NaOH \)[/tex] that is used is [tex]\( 0.50 M \)[/tex].

Mark the answer: [tex]\( 0.50 M \)[/tex]

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