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Chapter 16: Problem 9

Calculate \(\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]\) and \(\left[\mathrm{OH}^{-}\right]\) for each solution:(a) \(0.00165 \mathrm{M} \mathrm{HNO}_{3} ;\) (b) \(0.0087 \mathrm{M} \mathrm{KOH} ;\) (c) \(0.00213 \mathrm{M}\) \(\operatorname{Sr}(\mathrm{OH})_{2} ;(\mathrm{d}) 5.8 \times 10^{-4} \mathrm{M} \mathrm{HI}\)

Short Answer

Expert verified
\(\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]\) and \(\left[\mathrm{OH}^{-}\right]\) concentrations for each soultion are: (a) \(0.00165 M\) and \(6.06 \times 10^{-12} M\); (b) \(6.06 \times 10^{-12} M\) and \(0.0087 M\); (c) \(2.35 \times 10^{-12} M\) and \(0.00426 M\); (d) \(5.8 \times 10^{-4} M\) and \(1.72 \times 10^{-11} M\).

Step by step solution

01

Solve for (a): 0.00165 M HNO3

This is a strong acid. For every mole of HNO3, one mole of \(\mathrm{H}_{3} \mathrm{O}^{+}\) is produced. So, the concentration of \(\mathrm{H}_{3} \mathrm{O}^{+}\) ions is \(0.00165 M\). Knowing that \(KW=[\mathrm{H}_{3}\mathrm{O}^{+}][\mathrm{OH}^{-}]=1.0 \times 10^{-14} M^{2} \), we can find the concentration of \(\mathrm{OH}^{-}\) ions to be \( \frac{1.0 \times 10^{-14} M^2}{0.00165M} = 6.06 \times 10^{-12} M\).
02

Solve for (b): 0.0087 M KOH

KOH is a strong base. One mole of KOH results in one mole of \(\mathrm{OH}^{-}\) ions. So, the concentration of \(\mathrm{OH}^{-}\) is \(0.0087M\). Using the water ion product, we get the \(\mathrm{H}_{3} \mathrm{O}^{+}\) concentration to be \( \frac{1.0 \times 10^{-14} M^2}{0.0087M} = 1.15 \times 10^{-12} M \).
03

Solve for (c): 0.00213 M Sr(OH)2

This is a strong base with two hydroxyl ions for each formula unit. This means we have \(0.00213 M \times 2 = 0.00426 M\) of \(\mathrm{OH}^{-} \). The \(\mathrm{H}_{3} \mathrm{O}^{+}\) concentration is thus \( \frac{1.0 \times 10^{-14} M^2}{0.00426M} = 2.35 \times 10^{-12} M \).
04

Solve for (d): 5.8 x 10^{-4} M HI

HI is a strong acid. For each mole of HI, one mole of \(\mathrm{H}_{3} \mathrm{O}^{+}\) ions is produced. This makes the concentration of \(\mathrm{H}_{3} \mathrm{O}^{+}\) to be \( 5.8 \times 10^{-4} M \). Now, compute for the \(\mathrm{OH}^{-}\) concentration giving \( \frac{1.0 \times 10^{-14} M^2}{5.8 \times 10^{-4}M} = 1.72 \times 10^{-11} M \).

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