Chapter 16, Problem 46E

(a)

Interpretation Introduction

Interpretation:The concentration of ${\text{OH}}^{-}$ ions and $\text{pH}$ for $0.182\text{ M}$ $\text{KOH}$ should be calculated.

Concept Introduction: The $\text{pOH}$ is calculated by the formula,

  $\text{pOH}=-\log \left[{\text{OH}}^{-}\right]$

Where,

• $\left[{\text{OH}}^{-}\right]$ is the concentration of ${\text{OH}}^{-}$ ions. And, $\text{pOH+pH}=14$

(b)

Interpretation Introduction

Interpretation: The concentration of ${\text{OH}}^{-}$ ion and $\text{pH}$ of $3.165\text{ g}$

$\text{KOH}$ in $500.0\text{ mL}$ of solution should be calculated.

Concept Introduction: The $\text{pOH}$ is calculated by the formula,

  $\text{pOH}=-\log \left[{\text{OH}}^{-}\right]$

Where,

• $\left[{\text{OH}}^{-}\right]$ is the concentration of ${\text{OH}}^{-}$ions. And, $\text{pOH+pH}=14$

(c)

Interpretation Introduction

Interpretation:The concentration of ${\text{OH}}^{-}$ ions and $\text{pH}$ of $10.0\text{ mL}$ of $0.0105\text{ M}$

$\text{Ca}{\left(\text{OH}\right)}_2$ solution when it is diluted to $500.0\text{ mL}$ should be calculated.

Concept Introduction: The $\text{pOH}$ is calculated by the formula,

  $\text{pOH}=-\log \left[{\text{OH}}^{-}\right]$

Where,

• $\left[{\text{OH}}^{-}\right]$ is the concentration of ${\text{OH}}^{-}$ ions. And, $\text{pOH+pH}=14$

(d)

Interpretation Introduction

Interpretation: The concentration of ${\text{OH}}^{-}$ ions and $\text{pH}$ of a solution that if formed by mixing $20.0\text{ mL}$ of $0.015\text{ M Ba}{\left(\text{OH}\right)}_2$ solution to $40.0\text{ mL}$ of $8.2\times {10}^{-3}\text{ M NaOH}$ solution should be calculated.

Concept Introduction: The $\text{pOH}$ is calculated by the formula,

  $\text{pOH}=-\log \left[{\text{OH}}^{-}\right]$

Where,

• $\left[{\text{OH}}^{-}\right]$ is the concentration of ${\text{OH}}^{-}$ ions. And, $\text{pOH+pH}=14$