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Chapter 16: Problem 18

Calculate the \(\mathrm{pH}\) of water at \(40^{\circ} \mathrm{C}\), given that \(K_{\mathrm{w}}\) is \(3.8 \times 10^{-14}\) at this temperature.

Short Answer

Expert verified
The pH of water at \(40^{\circ} \mathrm{C}\) is \(7.21\).

Step by step solution

01

Understand the Problem and Recognize the Formula

The \(K_{\mathrm{w}}\) indicates product of the molar concentration of H+ and OH- ions in water. For pure water, [H+] = [OH-]. So we can express the \(K_{\mathrm{w}}\) as \(K_{\mathrm{w}}\ =\ [H+]^2\). The task requires finding the pH, which is the negative logarithm of the [H+] concentration. Therefore it includes recognizing and using these formulas.
02

Solve for [H+]

First, solve the \(K_{\mathrm{w}}\ =\ [H+]^2\) for [H+]. This will require square rooting both sides of the equation. Thus, we get [H+] = \(\sqrt{K_{\mathrm{w}}}\), [H+] = \(\sqrt{3.8 \times 10^{-14}}\) and using the given value of \(K_{\mathrm{w}}\).
03

Calculate [H+]

Using the square root that we calculated in step 2, we calculate the [H+] concentration as: [H+] = \(6.1644 \times 10^{-8}\) M.
04

Calculate pH

Once we have the [H+], we can calculate the pH using the formula: pH = -log[H+]. Therefore, pH = -log(\(6.1644 \times 10^{-8}\)), and pH = \(7.21\) at \(40^{\circ} \mathrm{C}\).

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Most popular questions from this chapter

Hydrocyanic acid (HCN) is a weak acid and a deadly poisonous compound that, in the gaseous form (hydrogen cyanide), is used in gas chambers. Why is it dangerous to treat sodium cyanide with acids (such as HCl) without proper ventilation?

HA and \(\mathrm{HB}\) are both weak acids although \(\mathrm{HB}\) is the stronger of the two. Will it take more volume of a \(0.10 M \mathrm{NaOH}\) solution to neutralize \(50.0 \mathrm{~mL}\) of \(0.10 \mathrm{M}\) HB than \(50.0 \mathrm{~mL}\) of \(0.10 \mathrm{M}\) HA?

What are the Lewis definitions of an acid and a base? In what way are they more general than the Bronsted definitions?

A \(0.400 M\) formic acid (HCOOH) solution freezes at \(-0.758^{\circ} \mathrm{C} .\) Calculate the \(K_{\mathrm{a}}\) of the acid at that temperature. (Hint: Assume that molarity is equal to molality. Carry your calculations to three significant figures and round off to two for \(K_{\mathrm{a}}\) .)

Identify the acid-base conjugate pairs in each of these reactions: (a) \(\mathrm{CH}_{3} \mathrm{COO}^{-}+\mathrm{HCN} \rightleftharpoons \mathrm{CH}_{3} \mathrm{COOH}+\mathrm{CN}^{-}\) (b) \(\mathrm{HCO}_{3}^{-}+\mathrm{HCO}_{3}^{-} \rightleftharpoons \mathrm{H}_{2} \mathrm{CO}_{3}+\mathrm{CO}_{3}^{2-}\) (c) \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}+\mathrm{NH}_{3} \rightleftharpoons \mathrm{HPO}_{4}^{2-}+\mathrm{NH}_{4}^{+}\) (d) \(\mathrm{HClO}+\mathrm{CH}_{3} \mathrm{NH}_{2} \rightleftharpoons \mathrm{CH}_{3} \mathrm{NH}_{3}^{+}+\mathrm{ClO}^{-}\) (e) \(\mathrm{CO}_{3}^{2-}+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{HCO}_{3}^{-}+\mathrm{OH}^{-}\) (f) \(\mathrm{CH}_{3} \mathrm{COO}^{-}+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{CH}_{3} \mathrm{COOH}+\mathrm{OH}^{-}\)
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