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Chapter 16: Problem 14

What is the \(\mathrm{pH}\) of the solution obtained when \(125 \mathrm{mL}\) of \(0.606 \mathrm{M} \mathrm{NaOH}\) is diluted to \(15.0 \mathrm{L}\) with water?

Short Answer

Expert verified
The pH of the solution, when \(125 mL\) of \(0.606 M NaOH\) is diluted to \(15.0 L\) with water, can be calculated using the steps outlined above.

Step by step solution

01

Calculate the new concentration of the NaOH solution

First, remember that when a solution is diluted, the amount of solute doesn't change, but the volume of the solvent increases. We can apply this principle in the formula \(M1V1 = M2V2\). Here, \(M1\) (original molarity) is \(0.606M\), \(V1\) (original volume) is \(125mL\) or \(0.125L\) and \(V2\) (new volume) is \(15.0L\). Solve for \(M2\) (new molarity), which represents the concentration of the NaOH solution after dilution.
02

Determine the concentration of OH- ions

NaOH, being a strong base, will completely dissociate in water to form Na+ and OH-. Hence the concentration of OH- ions will be the same as the concentration of the NaOH solution found in Step 1.
03

Calculate the pOH of the solution

The pOH of a solution can be calculated using the formula \(-\log[OH^-]\). Substitute the concentration of OH- ions found in Step 2 into this formula.
04

Calculate the pH of the solution

The pH and the pOH of a solution are related by the formula \(pH + pOH = 14\). Substitute the pOH found in Step 3 into this formula and solve for pH, which refers to the pH of the solution after dilution.

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Most popular questions from this chapter

\(50.00 \mathrm{mL}\) of \(0.0155 \mathrm{M} \mathrm{HI}(\mathrm{aq})\) is mixed with \(75.00 \mathrm{mL}\) of 0.0106 M KOH(aq). What is the pH of the final solution?

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