Chapter 24, Problem 7E

(a)

Interpretation Introduction

Interpretation:The concentration of ${\text{OH}}^{-}$ ions and $\text{pH}$ for $1.5\times {10}^{-3}\text{ M}$

$\text{Sr}{\left(\text{OH}\right)}_{\text{2}}$ 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 $2.250\text{ g}$

$\text{LiOH}$ in $250.0\text{ mL}$ 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 $1.00\text{ mL}$ of $0.175\text{ M}$

$\text{NaOH}$ solution when it is diluted to $2.00\text{ L}$ 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 $5.00\text{ mL}$ of $0.105\text{ M KOH}$ solution to $15.0\text{ mL}$ of $9.5\times {10}^{-2}\text{ M Ca}{\left(\text{OH}\right)}_2$ 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,

  $\text{pOH+pH}=14$