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Chapter 15: Problem 75

What volume of 0.100 \(\mathrm{M} \mathrm{NaOH}\) is required to neutralize 25.00 \(\mathrm{mL}\) of 0.110 \(\mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4} ?\)

Short Answer

Expert verified
The volume of 0.100 M \(NaOH\) required to neutralize 25.00 ml of 0.110 M \(H_{2}SO_{4}\) is 55 mL.

Step by step solution

01

Write the balanced equation for the reaction

The balanced chemical equation for the reaction is: \(H_{2} SO_{4} + 2 NaOH \rightarrow Na_{2}SO_{4} + 2H_{2}O\)
02

Calculate the moles of \(H_{2}SO_{4}\)

The molarity (M) equation is M= mol / L. Therefore, moles of \(H_{2}SO_{4}\) = Molarity * Volume = 0.110M * 0.0250L = 0.00275 moles
03

Find the moles of \(NaOH\)

From the balanced equation, it is evident that 1 mol of \(H_{2}SO_{4}\) reacts with 2 mol of \(NaOH\). Therefore, moles of \(NaOH\) = 2 * moles of \(H_{2}SO_{4}\) = 2 * 0.00275 moles = 0.0055 moles
04

Calculate volume of \(NaOH\)

Now we can use the molarity equation to find the volume of \(NaOH\).Volume = moles / Molarity = 0.0055 moles / 0.100 M = 0.055 L
05

Convert volume to milliliters

Since the given volume was provided in milliliters, we should convert the volume of \(NaOH\) to milliliters: 1 L = 1000 ml. Therefore, 0.055 L = 55 ml

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