Answer
6(a)
$$[OH^-] = 2.2 \times 10^{-14} \space M$$
7(b)
$$[OH^-] =9.5 \times 10^{-13} \space M$$
9(c)
$$[OH^-] =3.0 \times 10^{-14} \space M$$
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All these concentrations are lower than $[OH^-] = 1.0 \times 10^{-7}$, which makes sense, since they are all solutions made of acidic compounds, and therefore, acidic solutions.
Work Step by Step
6(a)
HBr is a strong acid: $[H_3O^+] = 0.45 M$
$$[OH^-] = \frac{1.0 \times 10^{-14}}{[H_3O^+]} = \frac{1.0 \times 10^{-14}}{ 0.45 } = 2.2 \times 10^{-14} \space M$$
7(b)
According to the calculation we made: $[H_3O^+] = 0.0105 M$
$$[OH^-] = \frac{1.0 \times 10^{-14}}{[H_3O^+]} = \frac{1.0 \times 10^{-14}}{ 0.0105 } = 9.5 \times 10^{-13} \space M$$
9(c)
According to the calculation we made: $[H_3O^+] = 0.336 M$
$$[OH^-] = \frac{1.0 \times 10^{-14}}{[H_3O^+]} = \frac{1.0 \times 10^{-14}}{ 0.336 } = 3.0 \times 10^{-14} \space M$$
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All these concentrations are lower than $[OH^-] = 1.0 \times 10^{-7}$, which makes sense, since they are all solutions made of acidic compounds, and therefore, acidic solutions.
