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Chapter 16: Problem 106

What is the \(\mathrm{pH}\) of a solution that is \(2.5 \times 10^{-9} \mathrm{M}\) in \(\mathrm{NaOH}\) ? Does your answer make sense? What assumption do we normally make that is not valid in this case?

Short Answer

Expert verified
The pH of the solution is approximately 5.4, which paradoxically suggests an acidic solution even though NaOH is a strong base. The result does not make sense because the low concentration of NaOH given makes the assumption of complete dissociation invalid in this case. In such dilute solutions, the self-ionization of water is significant and cannot be ignored, resulting in an acidic pH instead of the expected basic pH.

Step by step solution

01

Determine the concentration of hydroxide ions (OH⁻)

Given the concentration of NaOH is \(2.5\times10^{-9} \mathrm{M}\), and NaOH dissociates completely to yield OH⁻ ions, the concentration of OH⁻ ions in the solution is also \(2.5\times10^{-9} \mathrm{M}\).
02

Calculate the concentration of hydrogen ions (H⁺) using the ion product constant of water (\(K_w\))

Ion product constant of water, \(K_w\), is given by the relationship \(K_w = [\mathrm{H}⁺][\mathrm{OH}⁻]\). At 25°C, \(K_w = 1\times10^{-14}\). We can use this to calculate the H⁺ concentration: \[ [\mathrm{H}⁺] = \frac{K_w}{[\mathrm{OH}⁻]} \] Inserting the given values: \[ [\mathrm{H}⁺] = \frac{1 \times 10^{-14}}{2.5 \times 10^{-9}} \]
03

Solve for H⁺ ion concentration

Calculate the H⁺ ion concentration: \[ [\mathrm{H}⁺] = 4 \times 10^{-6} \mathrm{M} \]
04

Calculate the pH of the solution

Now that we have the H⁺ concentration, we can calculate the pH using the definition: \[ \mathrm{pH} = -\log_{10}[\mathrm{H}^+] \] Plug in the value of H⁺ concentration: \[ \mathrm{pH} = -\log_{10}(4 \times 10^{-6}) \]
05

Solve for pH

Calculate the pH of the solution: \[ \mathrm{pH} \approx 5.4 \]
06

Discuss the validity of the result and assumptions

The pH of the NaOH solution comes out to be around 5.4, which suggests an acidic solution. However, we know that NaOH is a strong base, so the answer does not make sense. This is because the concentration of NaOH given is very low, and the assumption that NaOH dissociates completely is not valid in this case. In such dilute solutions, the self-ionization of water becomes significant and cannot be ignored. The amount of H⁺ ions contributed by water is higher compared to the amount of OH⁻ ions contributed by the dissociation of NaOH in this dilute solution. This results in an acidic pH instead of the expected basic pH.

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Most popular questions from this chapter

Calculate \(\left[\mathrm{OH}^{-}\right]\) and \(\mathrm{pH}\) for each of the following strong base solutions: \((\mathbf{a}) 0.182 \mathrm{M} \mathrm{KOH},(\mathbf{b}) 3.165 \mathrm{g}\) of \(\mathrm{KOH}\) in 500.0 mL of solution, ( c ) 10.0 \(\mathrm{mL}\) of 0.0105 \(\mathrm{MCa}(\mathrm{OH})_{2}\) diluted to \(500.0 \mathrm{mL},(\mathbf{d})\) a solution formed by mixing 20.0 \(\mathrm{mL}\) of 0.015 \(M \mathrm{Ba}(\mathrm{OH})_{2}\) with 40.0 \(\mathrm{mL}\) of \(8.2 \times 10^{-3} \mathrm{M} \mathrm{NaOH}.\)

Predict the products of the following acid-base reactions, and predict whether the equilibrium lies to the left or to the right of the reaction arrow: (a) \(\mathrm{O}^{2-}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons\) (b) \(\mathrm{CH}_{3} \mathrm{COOH}(a q)+\mathrm{HS}^{-}(a q)\) (c) \(\mathrm{NO}_{2}^{-}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons\)

Benzoic acid \(\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COOH}\right)\) and aniline \(\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}\right)\) are both derivatives of benzene. Benzoic acid is an acid with \(K_{a}=6.3 \times 10^{-5}\) and aniline is a base with \(K_{a}=4.3 \times 10^{-10} .\) (a) What are the conjugate base of benzoic acid and the conjugate acid of aniline? (b) Anilinium chloride \(\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{3} \mathrm{Cl}\right)\) is a strong electrolyte that dissociates into anilinium ions \(\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{3}^{+}\right)\) and chloride ions. Which will be more acidic, a 0.10\(M\) solution of benzoic acid or a 0.10 M solution of anilinium chloride? (c) What is the value of the equilibrium constant for the following equilibrium? $$\begin{array}{c}{\mathrm{C}_{6} \mathrm{H}_{5} \operatorname{COOH}(a q)+\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}(a q) \rightleftharpoons} \\\ \quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad {\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COO}^{-}(a q)+\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{3}^{+}(a q)}\end{array}$$

The \(\mathrm{p} K_{\mathrm{b}}\) of water is ______ . (a) 1 (b) 7 (c) 14 (d) not defined (e) none of the above

The volume of an adult's stomach ranges from about 50 mL when empty to 1 when full. If the stomach volume is 400 mL and its contents have a pH of \(2,\) how many moles of \(\mathrm{H}^{+}\) does the stomach contain? Assuming that all the \(\mathrm{H}^{+}\) comes from \(\mathrm{HCl}\) , how many grams of sodium hydrogen carbonate will totally neutralize the stomach acid?
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