which equation correctly shows how you could calculate oh⁻ from h⁺?\noh⁻ = \\frac{1.0\\times10^{-14}}{h⁺}\nh⁺…

which equation correctly shows how you could calculate oh⁻ from h⁺?\noh⁻ = \\frac{1.0\\times10^{-14}}{h⁺}\nh⁺ = \\frac{1.0\\times10^{-14}}{oh⁻}

which equation correctly shows how you could calculate oh⁻ from h⁺?\noh⁻ = \\frac{1.0\\times10^{-14}}{h⁺}\nh⁺ = \\frac{1.0\\times10^{-14}}{oh⁻}

Answer

Explanation:

Step1: Recall ion - product constant of water

The ion - product constant of water $K_w=[H^+][OH^-]$, and at $25^{\circ}C$, $K_w = 1.0\times10^{-14}$.

Step2: Solve for $[OH^-]$

We can re - arrange the equation $K_w=[H^+][OH^-]$ for $[OH^-]$. Dividing both sides of the equation by $[H^+]$, we get $[OH^-]=\frac{K_w}{[H^+]}$. Since $K_w = 1.0\times10^{-14}$ at $25^{\circ}C$, then $[OH^-]=\frac{1.0\times10^{-14}}{[H^+]}$.

Answer:

$[OH^-]=\frac{1.0\times10^{-14}}{[H^+]}$ (the first option)