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2. How many liters of 4.5 M HCl are needed to neutralize 80 mL of a 1.5 M NaOH
solution?

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Answer:

In this problem, we can use the concept of acid-base titration to determine the volume of HCl required to neutralize the NaOH solution. The balanced chemical equation for the reaction between HCl and NaOH is:

HCl + NaOH → NaCl + H2O

From the equation, we can see that 1 mole of HCl reacts with 1 mole of NaOH. Therefore, the number of moles of NaOH in the solution is:

moles of NaOH = Molarity × Volume = 1.5 M × 0.080 L = 0.12 moles

Since 1 mole of HCl reacts with 1 mole of NaOH, the number of moles of HCl required to neutralize the NaOH solution is also 0.12 moles.

The molarity of the HCl solution is 4.5 M. We can use the following equation to calculate the volume of HCl required:

moles of HCl = Molarity × Volume

Rearranging the equation gives:

Volume = moles of HCl / Molarity

Substituting the values we have:

Volume = 0.12 moles / 4.5 M = 0.0267 L

Therefore, the volume of 4.5 M HCl solution required to neutralize the 1.5 M NaOH solution is 0.0267 L, or 26.7 mL (since 1 L = 1000 mL).

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