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Chapter 16: Problem 121

Calculate the number of \(\mathrm{H}^{+}(a q)\) ions in \(1.0 \mathrm{~mL}\) of pure water at \(25^{\circ} \mathrm{C}\).

Short Answer

Expert verified
There are approximately \(6.022 \times 10^{13}\) H+ ions in 1.0 mL of pure water at 25掳C.

Step by step solution

01

1. Find the ion product constant (Kw) at 25掳C

At 25掳C, the ion product constant for water is given as: \(K_w = [H^+][OH^-] = 1.0 \times 10^{-14}\)
02

2. Calculate the concentration of H+ ions in pure water

In pure water, the concentrations of H+ and OH- ions are equal, as the number of H+ ions formed is equal to the number of OH- ions. Therefore: \([H^+] = [OH^-]\) We can then rewrite Kw as: \[K_w = [H^+]^2\] Now, solve for [H+]: \[[H^+] = \sqrt{K_w} = \sqrt{1.0 \times 10^{-14}}\]
03

3. Determine the concentration of H+ ions

Calculating the square root: \[[H^+] = 1.0 \times 10^{-7} \,M\]
04

4. Calculate the number of H+ ions in 1.0 mL of pure water

To do this, we will first convert the volume from mL to L: \[1.0\, mL = 1.0 \times 10^{-3} L\] Now, we can use the concentration formula (M = mol/L) to determine the number of moles of H+ ions in the given volume: \[moles\,of\,H^+ = [H^+] \times volume\] \[moles\,of\,H^+ = (1.0 \times 10^{-7}\,M) \times (1.0 \times 10^{-3}\,L)\] Calculating the product: \[moles\,of\,H^+ = 1.0 \times 10^{-10}\,mol\]
05

5. Determine the number of H+ ions

Finally, multiply the number of moles with Avogadro's number to find the total number of H+ ions: \[number\,of\,H^+ = moles\,of\,H^+ \times Avogadro's\,number\] \[number\,of\,H^+ = (1.0 \times 10^{-10}\,mol) \times (6.022 \times 10^{23}\,ion/mol)\] Calculating the product: \[number\,of\,H^+ \approx 6.022 \times 10^{13}\,ion\] There are approximately \(6.022 \times 10^{13}\) H+ ions in 1.0 mL of pure water at 25掳C.

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Most popular questions from this chapter

Calculate \(\left[\mathrm{H}^{+}\right]\) for each of the following solutions, and indicate whether the solution is acidic, basic, or neutral: (a) \(\left[\mathrm{OH}^{-}\right]=0.00045 M ;\) (b) \(\left[\mathrm{OH}^{-}\right]=8.8 \times 10^{-9} \mathrm{M} ;(\mathrm{c}) \mathrm{a}\) solution in which \(\left[\mathrm{OH}^{-}\right]\) is 100 times greater than \(\left[\mathrm{H}^{+}\right]\).

Indicate whether each of the following statements is correct or incorrect. For those that are incorrect, explain why they are wrong. (a) Every Br酶nsted-Lowry acid is also a Lewis acid. (b) Every Lewis acid is also a Br酶nsted-Lowry acid. (c) Conjugate acids of weak bases produce more acidic solutions than conjugate acids of strong bases. (d) \(\mathrm{K}^{+}\) ion is acidic in water because it causes hydrating water molecules to become more acidic. (e) The percent ionization of a weak acid in water increases as the concentration of acid decreases.

Which member of each pair produces the more acidic aqueous solution: (a) \(\mathrm{ZnBr}_{2}\) or \(\mathrm{CdCl}_{2}\), (b) \(\mathrm{CuCl}\) or \(\mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2},(\mathrm{c}) \mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2}\) or \(\mathrm{NiBr}_{2} ?\) Explain.

A \(0.100 \mathrm{M}\) solution of bromoacetic acid \(\left(\mathrm{BrCH}_{2} \mathrm{COOH}\right)\) is \(13.2 \%\) ionized. Calculate \(\left[\mathrm{H}^{+}\right],\left[\mathrm{BrCH}_{2} \mathrm{COO}^{-}\right]\), and \(\left[\mathrm{BrCH}_{2} \mathrm{COOH}\right]\).

Identify the Lewis acid and Lewis base among the reactants in each of the following reactions: (a) \(\mathrm{Fe}\left(\mathrm{ClO}_{4}\right)_{3}(s)+6 \mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons\) \(\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}{ }^{3+}(a q)+3 \mathrm{ClO}_{4}^{-}(a q)\) (b) \(\mathrm{CN}^{-}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{HCN}(a q)+\mathrm{OH}^{-}(a q)\) (c) \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{~N}(g)+\mathrm{BF}_{3}(g) \rightleftharpoons\left(\mathrm{CH}_{3}\right)_{3} \mathrm{NBF}_{3}(s)\) (d) \(\mathrm{HIO}(l q)+\mathrm{NH}_{2}^{-}(l q) \rightleftharpoons \mathrm{NH}_{3}(l q)+\mathrm{IO}^{-}(l q)\) (lq denotes liquid ammonia as solvent)
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