1. the poh of a solution is 2.35. what is its hydroxide ion concentration (oh⁻) in mol/l?\n2. the poh of a…

1. the poh of a solution is 2.35. what is its hydroxide ion concentration (oh⁻) in mol/l?\n2. the poh of a solution is 0.46. what is its hydroxide ion concentration (oh⁻) in mol/l?\n3. the poh of a solution is 10.96. what is its hydrogen ion concentration (h⁺) in mol/l?\n4. the poh of a solution is 8.18. what is its hydrogen ion concentration (h⁺) in mol/l?\n5. the poh of a solution is 11.1. what is its hydroxide ion concentration (oh⁻) in mol/l?\n6. the hydrogen ion concentration (h⁺) of a solution is 3.02×10⁻⁷ mol/l. what is its poh?\n7. the poh of a solution is 8.68. what is its hydroxide ion concentration (oh⁻) in mol/l?\n8. the hydroxide ion concentration (oh⁻) of a solution is 8.3×10⁻⁵ mol/l. what is its poh?\n9. the poh of a solution is 9.72. what is its hydroxide ion concentration (oh⁻) in mol/l?

1. the poh of a solution is 2.35. what is its hydroxide ion concentration (oh⁻) in mol/l?\n2. the poh of a solution is 0.46. what is its hydroxide ion concentration (oh⁻) in mol/l?\n3. the poh of a solution is 10.96. what is its hydrogen ion concentration (h⁺) in mol/l?\n4. the poh of a solution is 8.18. what is its hydrogen ion concentration (h⁺) in mol/l?\n5. the poh of a solution is 11.1. what is its hydroxide ion concentration (oh⁻) in mol/l?\n6. the hydrogen ion concentration (h⁺) of a solution is 3.02×10⁻⁷ mol/l. what is its poh?\n7. the poh of a solution is 8.68. what is its hydroxide ion concentration (oh⁻) in mol/l?\n8. the hydroxide ion concentration (oh⁻) of a solution is 8.3×10⁻⁵ mol/l. what is its poh?\n9. the poh of a solution is 9.72. what is its hydroxide ion concentration (oh⁻) in mol/l?

Answer

Explanation:

Step1: Recall the formula

The relationship between pOH and hydroxide - ion concentration is $[\text{OH}^-]=10^{-\text{pOH}}$. Also, we know that $[\text{H}^+][\text{OH}^-]=10^{- 14}$, so $[\text{H}^+]=\frac{10^{-14}}{[\text{OH}^-]}$.

Step2: Solve for question 1

Given $\text{pOH} = 2.35$, then $[\text{OH}^-]=10^{-2.35}\approx4.47\times 10^{-3}\text{ mol/L}$.

Step3: Solve for question 2

Given $\text{pOH}=0.46$, then $[\text{OH}^-]=10^{-0.46}\approx0.35\text{ mol/L}$.

Step4: Solve for question 3

First, find $[\text{OH}^-]$ using $[\text{OH}^-]=10^{-\text{pOH}}$. Since $\text{pOH} = 10.96$, $[\text{OH}^-]=10^{-10.96}$. Then, use $[\text{H}^+]=\frac{10^{-14}}{[\text{OH}^-]}=\frac{10^{-14}}{10^{-10.96}}=10^{-3.04}\approx9.12\times 10^{-4}\text{ mol/L}$.

Step5: Solve for question 4

First, find $[\text{OH}^-]$ using $[\text{OH}^-]=10^{-\text{pOH}}$. Since $\text{pOH}=8.18$, $[\text{OH}^-]=10^{-8.18}$. Then, $[\text{H}^+]=\frac{10^{-14}}{[\text{OH}^-]}=\frac{10^{-14}}{10^{-8.18}}=10^{-5.82}\approx1.51\times 10^{-6}\text{ mol/L}$.

Step6: Solve for question 5

Given $\text{pOH}=11.1$, then $[\text{OH}^-]=10^{-11.1}\approx7.94\times 10^{-12}\text{ mol/L}$.

Step7: Solve for question 6

First, find $[\text{OH}^-]$ using $[\text{H}^+][\text{OH}^-]=10^{-14}$. So $[\text{OH}^-]=\frac{10^{-14}}{[\text{H}^+]}$, with $[\text{H}^+]=3.02\times 10^{-7}\text{ mol/L}$, then $[\text{OH}^-]=\frac{10^{-14}}{3.02\times 10^{-7}}\approx3.31\times 10^{-8}\text{ mol/L}$. And $\text{pOH}=-\log_{10}[\text{OH}^-]=-\log_{10}(3.31\times 10^{-8})\approx7.48$.

Step8: Solve for question 7

Given $\text{pOH}=8.68$, then $[\text{OH}^-]=10^{-8.68}\approx2.09\times 10^{-9}\text{ mol/L}$.

Step9: Solve for question 8

Given $[\text{OH}^-]=8.3\times 10^{-5}\text{ mol/L}$, then $\text{pOH}=-\log_{10}(8.3\times 10^{-5})\approx4.08$.

Step10: Solve for question 9

Given $\text{pOH}=9.72$, then $[\text{OH}^-]=10^{-9.72}\approx1.90\times 10^{-10}\text{ mol/L}$.

Answer:

  1. $4.47\times 10^{-3}\text{ mol/L}$
  2. $0.35\text{ mol/L}$
  3. $9.12\times 10^{-4}\text{ mol/L}$
  4. $1.51\times 10^{-6}\text{ mol/L}$
  5. $7.94\times 10^{-12}\text{ mol/L}$
  6. $7.48$
  7. $2.09\times 10^{-9}\text{ mol/L}$
  8. $4.08$
  9. $1.90\times 10^{-10}\text{ mol/L}$