1. calculate the h⁺, oh⁻, ph and poh for 0.75 m nh₄cl.

Explanation:

Step1: Consider hydrolysis of NH₄Cl

NH₄⁺ + H₂O ⇌ NH₃·H₂O + H⁺. Let $x$ be the concentration of H⁺ ions at equilibrium. Then the equilibrium concentrations are [NH₄⁺] = 0.75 - $x$, [NH₃·H₂O] = $x$, [H⁺] = $x$. The $K_a$ for NH₄⁺ is $\frac{K_w}{K_b}$, where $K_b$ for NH₃ is $1.8\times10^{-5}$ and $K_w = 1.0\times10^{-14}$, so $K_a=\frac{1.0\times10^{-14}}{1.8\times10^{-5}}\approx5.6\times10^{-10}$. Since $K_a$ is very small, 0.75 - $x\approx0.75$.

Step2: Set up the $K_a$ expression

$K_a=\frac{[NH_3\cdot H_2O][H^+]}{[NH_4^+]}$, so $5.6\times10^{-10}=\frac{x\cdot x}{0.75}$.

Step3: Solve for $x$ (i.e., [H⁺])

$x^2 = 5.6\times10^{-10}\times0.75$, $x=\sqrt{5.6\times10^{-10}\times0.75}\approx2.05\times10^{-5}$ M. So $[H^+]=2.05\times10^{-5}$ M.

Step4: Calculate [OH⁻]

Using $[H^+][OH^-]=K_w$, $[OH^-]=\frac{K_w}{[H^+]}=\frac{1.0\times10^{-14}}{2.05\times10^{-5}}\approx4.88\times10^{-10}$ M.

Step5: Calculate pH

$pH = -\log[H^+]=-\log(2.05\times10^{-5})\approx4.69$.

Step6: Calculate pOH

$pOH=14 - pH = 14 - 4.69 = 9.31$.

Answer:

$[H^+]=2.05\times10^{-5}$ M, $[OH^-]=4.88\times10^{-10}$ M, $pH = 4.69$, $pOH = 9.31$