rank the following solutions from the highest h3o+ (1) to the lowest h3o+ (5). h3o+=3.16×10−4 m…

rank the following solutions from the highest h3o+ (1) to the lowest h3o+ (5). h3o+=3.16×10−4 m oh−=4.35×10−2 m ph = 1.05 poh = 7.0 poh = 4.0 done

rank the following solutions from the highest h3o+ (1) to the lowest h3o+ (5). h3o+=3.16×10−4 m oh−=4.35×10−2 m ph = 1.05 poh = 7.0 poh = 4.0 done

Answer

Answer:

  1. $\text{pH}=1.05$
  2. $[\text{H}_3\text{O}^+]=3.16\times 10^{-4}\text{ M}$
  3. $\text{pOH}=7.0$
  4. $\text{pOH}=4.0$
  5. $[\text{OH}^-]=4.35\times 10^{-2}\text{ M}$

Explanation:

Step1: Calculate $[\text{H}_3\text{O}^+]$ for $\text{pH} = 1.05$

$[\text{H}_3\text{O}^+]=10^{-\text{pH}}=10^{- 1.05}\approx8.91\times 10^{-2}\text{ M}$

Step2: Calculate $[\text{H}_3\text{O}^+]$ for $[\text{OH}^-]=4.35\times 10^{-2}\text{ M}$

Use $K_w=[\text{H}_3\text{O}^+][\text{OH}^-]=1.0\times 10^{-14}$. So $[\text{H}_3\text{O}^+]=\frac{K_w}{[\text{OH}^-]}=\frac{1.0\times 10^{-14}}{4.35\times 10^{-2}}\approx2.30\times 10^{-13}\text{ M}$

Step3: Calculate $[\text{H}_3\text{O}^+]$ for $\text{pOH}=7.0$

First, $\text{pH}=14 - \text{pOH}=14 - 7.0 = 7.0$. Then $[\text{H}_3\text{O}^+]=10^{-\text{pH}}=10^{-7}\text{ M}$

Step4: Calculate $[\text{H}_3\text{O}^+]$ for $\text{pOH}=4.0$

First, $\text{pH}=14-\text{pOH}=14 - 4.0=10.0$. Then $[\text{H}_3\text{O}^+]=10^{-\text{pH}}=10^{-10}\text{ M}$

Step5: Compare values

Comparing $8.91\times 10^{-2}\text{ M}$, $3.16\times 10^{-4}\text{ M}$, $10^{-7}\text{ M}$, $10^{-10}\text{ M}$, $2.30\times 10^{-13}\text{ M}$, we get the ranking.