rank the following solutions from the highest h₃o⁺ (1) to the lowest h₃o⁺ (5).\nh₃o⁺=3.16×10⁻⁴…

rank the following solutions from the highest h₃o⁺ (1) to the lowest h₃o⁺ (5).\nh₃o⁺=3.16×10⁻⁴ m\noh⁻=4.35×10⁻² m\nph = 1.05\npoh = 7.0\npoh = 4.0

rank the following solutions from the highest h₃o⁺ (1) to the lowest h₃o⁺ (5).\nh₃o⁺=3.16×10⁻⁴ m\noh⁻=4.35×10⁻² m\nph = 1.05\npoh = 7.0\npoh = 4.0

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 given $[\text{OH}^-]$

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

Step2: Calculate $[\text{H}_3\text{O}^+]$ for given $\text{pH}$

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

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

First find $\text{pH}$ using $\text{pH}+\text{pOH}=14$. For $\text{pOH}=7.0$, $\text{pH}=14 - 7.0 = 7.0$, then $[\text{H}_3\text{O}^+]=10^{-7.0}=1.0\times 10^{-7}\text{ M}$. For $\text{pOH}=4.0$, $\text{pH}=14 - 4.0=10.0$, then $[\text{H}_3\text{O}^+]=10^{-10.0}=1.0\times 10^{-10}\text{ M}$.

Step4: Rank the $[\text{H}_3\text{O}^+]$ values

Compare all $[\text{H}_3\text{O}^+]$ values: $8.91\times 10^{-2}\text{ M}>3.16\times 10^{-4}\text{ M}>1.0\times 10^{-7}\text{ M}>1.0\times 10^{-10}\text{ M}>2.30\times 10^{-13}\text{ M}$.