Answer
(a) $[H_3O^+] = 0.00165 M$ and $[OH^-] = 6.06 \times 10^{-12}M$
(b) $[H_3O^+] = 1.15 \times 10^{-12}M$ and $[OH^-] = 0.0087M$
(c) $[H_3O^+] =2.35 \times 10^{-12}M$ and $[OH^-] = 0.00426M$
(d) $[H_3O^+] = 5.8 \times 10^{-4}M$ and $[OH^-] = 1.72 \times 10^{-11}M$
Work Step by Step
For (a) and (d)
- Since we got strong acids, the concentration of the acid = $[H_3O^+]$.
- To calculate the $[OH^-]$, use the equation:
$[H_3O^+] * [OH^-] = 10^{-14}$
For (b)
- Since we got a strong base, the concentration of the base = $[OH^-]$.
- To calculate the $[H_3O^+]$, use the equation:
$[H_3O^+] * [OH^-] = 10^{-14}$
For (c)
- We got a strong base, but it has 2 $OH^-$ for each molecule, therefore $[OH^-]$ will be 2 times $0.00213M$.
- To calculate the $[H_3O^+]$, use the equation:
$[H_3O^+] * [OH^-] = 10^{-14}$
