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Imagine that you have 15 $\mathrm{mL}$ of a 0.025 $\mathrm{MCl}$ solution.a. How many moles of $\mathrm{H}^{+}$ are in the 15 $\mathrm{mL}$ ?b. If you add 25 $\mathrm{mL}$ of water, what is the new concentration of the solution?c. What is the final pH after adding 25 $\mathrm{mL}$ ?

a.$3.75 \times 40^{-4}$b.0.00938c.2,03

Chemistry 101

Chapter 16

Acidic Toxins

Section 4

Dilution

Intro to Chemistry

Atoms, Molecules and Ions

University of Maryland - University College

Brown University

University of Toronto

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question. Number eight is very similar to question number seven. It asks us to convert moles and volume. I'm sorry polarity in volume into moles and then conduct a dilution calculation. Let's start with part A In part A. We have 15 million liters of the solution. That is 0.25 Moeller. We convert the mill leaders into leaders and then the mole leaders into moles and recognize that for every one mole hcl, we have one mole hydro Nia, My on. So we get 3.75 times 10 to the negative. Four moles Hydro Nia, my on If we take the 3.75 times 10 The negative four moles Hydro Nia, My on that was in 15 mL, or 150.15 leaders and add to it another 25 mL or 250.25 leaders. Then the moles divided by the new volume will give us a new concentration. The 0.938 Moeller pH. Then it's just a negative log of this hydro Nehemiah and concentration or 203

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