Chapter 15: Problem 78

If 50.00 \(\mathrm{mL}\) of 1.000 \(\mathrm{M} \mathrm{HI}\) is neutralized by 35.41 \(\mathrm{mL}\) of \(\mathrm{KOH}\) , what is the molarity of the \(\mathrm{KOH}\) solution?

Short Answer

Expert verified

The molarity of the KOH solution is 1.412 M

Step by step solution

Write down the balanced chemical equation

The reaction of potassium hydroxide (KOH) with hydroiodic acid (HI) can be written as follows: \(\mathrm{HI} + \mathrm{KOH} \rightarrow \mathrm{H_2O} + \mathrm{KI}\)

Analyzing the stoichiometry

From the balanced equation, it can be observed that one mole of HI reacts with one mole of KOH to produce one mole of water and one mole of potassium iodide. It means the mole of HI and KOH is 1:1.

Calculate the moles of HI

Using the Molarity formula \(Molarity = Moles/Volume\), the number of moles can be found by multiplying the molarity with the volume (in liters). Here, the molarity \(M\) is 1.000 \(\mathrm{M}\) and the volume \(V\) is 50.00 mL or 0.05000 L. Therefore, the moles of HI \(n_{HI}\) = \(M * V = 1.000 * 0.050 = 0.050\) moles.

Find the molarity of KOH

Since the moles of HI and KOH are in 1:1 ratio, the moles of KOH will also be 0.050 moles. The volume of KOH is given as 35.41 mL or 0.03541 L. The molarity of KOH can now be calculated using the molarity formula again: \(M_{KOH} = n_{KOH} / V_{KOH} = 0.050 / 0.03541 = 1.412 M\).

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If 50.00 \(\mathrm{mL}\) of 1.000 \(\mathrm{M} \mathrm{HI}\) is neutralized by 35.41 \(\mathrm{mL}\) of \(\mathrm{KOH}\) , what is the molarity of the \(\mathrm{KOH}\) solution?

Short Answer

Step by step solution

Write down the balanced chemical equation

Analyzing the stoichiometry

Calculate the moles of HI

Find the molarity of KOH

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