Chapter 16 - Exercises - Page 772: 52

\begin{vmatrix} [H_3O^+] & [OH^-] & pH & Acidic \space or Basic \\ 3.5 \times 10^{-3} & 2.9 \times 10^{-12} & 2.46 & Acidic \\ 2.6 \times 10^{-8} & 3.8 \times 10^{-7} & 7.59 & Basic\\ 1.8 \times 10^{-9} & 5.6 \times 10^{-6} & 8.74 & Basic \\ 7.1 \times 10^{-8} & 1.4 \times 10^{-7} & 7.15 & Basic \end{vmatrix}

Work Step by Step

We find: a. $$[OH^-] = \frac{1.0 \times 10^{-14}}{[H_3O^+]} = \frac{1.0 \times 10^{-14}}{ 3.5 \times 10^{-3} } = 2.9 \times 10^{-12} \space M$$ $$pH = -log[H_3O^+] = -log( 3.5 \times 10^{-3} ) = 2.46$$ pH < 7: Acidic b. $$[H_3O^+] = \frac{1.0 \times 10^{-14}}{[OH^-]} = \frac{1.0 \times 10^{-14}}{ 3.8 \times 10^{-7} } = 2.6 \times 10^{-8} \space M$$ $$pH = -log[H_3O^+] = -log( 2.6 \times 10^{-8} ) = 7.59$$ pH > 7: Basic c. $$[OH^-] = \frac{1.0 \times 10^{-14}}{[H_3O^+]} = \frac{1.0 \times 10^{-14}}{ 1.8 \times 10^{-9} } = 5.6 \times 10^{-6} \space M$$ $$pH = -log[H_3O^+] = -log( 1.8 \times 10^{-9} ) = 8.74$$ pH > 7: Basic d. $$[H_3O^+] = 10^{-pH} = 10^{-7.15} = 7.1 \times 10^{-8} \space M$$ $$[OH^-] = \frac{1.0 \times 10^{-14}}{[H_3O^+]} = \frac{1.0 \times 10^{-14}}{ 7.1 \times 10^{-8} } = 1.4 \times 10^{-7} \space M$$ pH > 7: Basic

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