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Chapter 14: Problem 14

Why is the \(\mathrm{pH}\) of water at \(25^{\circ} \mathrm{C}\) equal to \(7.00\) ?

Short Answer

Expert verified
The pH of water at \(25^{\circ} \mathrm{C}\) is equal to 7.00 because the ion product constant for water (Kw) at this temperature is \(1 \times 10^{-14}\). In pure water, the concentrations of hydrogen ions [\(H^+\)] and hydroxide ions [\(OH^-\)] are equal. Thus, each ion concentration is \(1 \times 10^{-7}\) M. Using the pH formula, \(pH = -\log[H^+]\), we get a value of 7.00 for the pH of pure water at \(25^{\circ} \mathrm{C}\).

Step by step solution

01

Understand the definition of pH

The pH of a solution is defined as the negative logarithm of the hydrogen ion concentration: \[ pH = -\log[H^{+}] \] where \(H^{+}\) represents the concentration of hydrogen ions in the solution.
02

Calculate the ion product constant for water (Kw) at 25°C

In water, a small fraction of the water molecules dissociate into hydrogen ions (H⁺) and hydroxide ions (OH⁻). This can be represented by the following chemical equation: \[ H_{2}O \rightleftharpoons H^{+} + OH^{-} \] The ion product constant for water (Kw) is the equilibrium constant for this dissociation, and it is given by the product of the concentrations of hydrogen and hydroxide ions: \[ K_w = [H^{+}][OH^{-}] \] At 25°C, the value of Kw is equal to \(1 \times 10^{-14}\).
03

Equate the concentrations of hydrogen and hydroxide ions in pure water

In pure water, the concentrations of hydrogen ions and hydroxide ions are equal: \[ [H^{+}] = [OH^{-}] \] Since both concentrations are equal, we can introduce a variable x for both concentrations: \[ [H^{+}] = [OH^{-}] = x \]
04

Solve for the hydrogen ion concentration

Now, we can substitute this value of x into the equation for Kw to find the concentration of hydrogen ions in pure water at 25°C: \[ K_w = x^2 \] Solving for x, we get: \[ x = \sqrt{K_w} = \sqrt{1 \times 10^{-14}} \] Thus, \(x = 1 \times 10^{-7}\), which is the concentration of hydrogen ions in pure water at 25°C.
05

Calculate the pH of water at 25°C

Finally, we can use the definition of pH to calculate the pH of water at 25°C: \[ pH = -\log[H^{+}] = -\log(1 \times 10^{-7}) \] Therefore, the pH of water at 25°C is equal to 7.00.

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Most popular questions from this chapter

Classify each of the following as a strong acid, weak acid, strong base, or weak base in aqueous solution. a. \(\mathrm{HNO}_{2}\) b. \(\mathrm{HNO}_{3}\) c. \(\mathrm{CH}_{3} \mathrm{NH}_{2}\) d. \(\mathrm{NaOH}\) e. \(\mathrm{NH}_{3}\) f. \(\mathrm{HF}\) g. h. \(\mathrm{Ca}(\mathrm{OH})_{2}\) i. \(\mathrm{H}_{2} \mathrm{SO}_{4}\)

One mole of a weak acid HA was dissolved in \(2.0 \mathrm{~L}\) of solution. After the system had come to equilibrium, the concentration of HA was found to be \(0.45 M .\) Calculate \(K_{\mathrm{a}}\) for HA.

For the following, mix equal volumes of one solution from Group I with one solution from Group II to achieve the indicated \(\mathrm{pH}\). Calculate the \(\mathrm{pH}\) of each solution. Group I: \(0.20 \mathrm{M} \mathrm{NH}_{4} \mathrm{Cl}, 0.20 \mathrm{MHCl}, 0.20 \mathrm{M} \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{3} \mathrm{Cl}, 0.20\) \(M\left(\mathrm{C}_{2} \mathrm{H}_{5}\right)_{3} \mathrm{NHCl}\) Group II: \(0.20 \mathrm{M} \mathrm{KOI}, 0.20 \mathrm{M} \mathrm{NaCN}, 0.20 \mathrm{M} \mathrm{KOCl}, 0.20 \mathrm{M}\) \(\mathrm{NaNO}_{2}\) a. the solution with the lowest \(\mathrm{pH}\) b. the solution with the highest \(\mathrm{pH}\) c. the solution with the \(\mathrm{pH}\) closest to \(7.00\)

The \(\mathrm{pH}\) of a \(1.00 \times 10^{-2} \mathrm{M}\) solution of cyanic acid (HOCN) is \(2.77\) at \(25^{\circ} \mathrm{C}\). Calculate \(K_{\mathrm{a}}\) for \(\mathrm{HOCN}\) from this result.

Write out the stepwise \(K_{\mathrm{a}}\) reactions for the diprotic acid \(\mathrm{H}_{2} \mathrm{SO}_{3}\).
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