chem 12\n1. calculate the h⁺, oh⁻, ph and poh for 0.50 m h₃bo₃.

Explanation:

Step1: Write dissociation equation

$H_3BO_3 + H_2O\rightleftharpoons H_2BO_3^-+H_3O^+$ The $K_a$ of $H_3BO_3$ is $5.8\times 10^{-10}$. Let $[H^+]=x$, then $[H_2BO_3^-]=x$ and $[H_3BO_3]=0.50 - x\approx0.50$ (since $K_a$ is very small).

Step2: Use $K_a$ expression

$K_a=\frac{[H^+][H_2BO_3^-]}{[H_3BO_3]}$, so $5.8\times 10^{-10}=\frac{x\cdot x}{0.50}$ $x^2 = 5.8\times 10^{-10}\times0.50$ $x=\sqrt{5.8\times 10^{-10}\times0.50}\approx1.7\times 10^{-5}\ M$ So $[H^+]=1.7\times 10^{-5}\ M$

Step3: Calculate $[OH^-]$

We know that $K_w=[H^+][OH^-]=1.0\times 10^{-14}$ $[OH^-]=\frac{K_w}{[H^+]}=\frac{1.0\times 10^{-14}}{1.7\times 10^{-5}}\approx5.9\times 10^{-10}\ M$

Step4: Calculate pH

$pH =-\log[H^+]=-\log(1.7\times 10^{-5})\approx4.77$

Step5: Calculate pOH

$pOH=14 - pH=14 - 4.77 = 9.23$

Answer:

$[H^+]=1.7\times 10^{-5}\ M$, $[OH^-]=5.9\times 10^{-10}\ M$, $pH\approx4.77$, $pOH\approx9.23$