Chapter 12, Problem 12.31P

(a)

Interpretation Introduction

Interpretation:

Volume of $\text{0}{\text{.108 M H}}_2{\text{SO}}_{\text{4}}$ solution needed to neutralize $\text{15}\text{.0 μ}L$ solution of $\text{0}\text{.010 M Ca}{\left(\text{OH}\right)}_2$ has to be calculated.

Concept Introduction:

The dilution formula is given as follows:

  $n_1{M}_1{V}_1=n_2{M}_2{V}_2$

Here,

$M_1$ denotes the molarity of the solution before dilution.

$V_1$ denotes the volume of the solution before dilution.

$M_2$ denotes the molarity of solution after dilution.

$V_2$ denotes the volume of the solution after dilution.

$n_1$ and $n_2$ are n factors of acid and base respectively.

Answer to Problem 12.31P

Volume of $\text{0}{\text{.108 M H}}_2{\text{SO}}_{\text{4}}$ solution needed to neutralize $\text{15}\text{.0 μ}L$ solution of $\text{0}\text{.010 M Ca}{\left(\text{OH}\right)}_2$ is $0.1851\times {10}^{-6}\text{ L}$.

Explanation of Solution

The conversion factor to convert liters to is $\text{μL}$ as follows:

  $\text{1 μL}={\text{10}}^{-6}\text{ L}$

Hence convert $\text{15 μL}$ to $\text{L}$ as follows:

  $\begin{array}{c}\text{Volume}=\left(\text{15 μL}\right)\left(\frac{{\text{10}}^{-6}\text{ L}}{\text{1 μL}}\right)\\ =15\times {10}^{-6}\text{ L}\end{array}$

The dilution formula is given as follows:

  $n_1{M}_1{V}_1=n_2{M}_2{V}_2$        (1)

Substitute 1 for $n_1$, 2 for $n_2$, $\text{0}\text{.108 M}$ for $M_1$, $\text{0}\text{.010 M}$ for $M_2$, $15\times {10}^{-6}\text{ L}$ for $V_2$ in equation (1).

  $\left(1\right)\left(0.108\text{ M}\right)\left(V_1\right)=\left(2\right)\left(\text{0}\text{.010 M}\right)\left(15\times {10}^{-6}\text{ L}\right)$        (2)

Rearrange equation (2) to calculate the value of $V_1$.

  $\begin{array}{c}{V}_1=\frac{\left(2\right)\left(\text{0}\text{.010 M}\right)\left(15\times {10}^{-6}\text{ L}\right)}{\left(1\right)\left(0.108\text{ M}\right)}\\ =0.1851\times {10}^{-6}\text{ L}\end{array}$

(b)

Interpretation Introduction

Interpretation:

Volume of $\text{0}{\text{.300 M H}}_2{\text{SO}}_{\text{4}}$ solution needed to neutralize $\text{25}\text{.0 }\mathrm{mL}$ solution of $\text{0}\text{.200 M NaOH}$ has to be calculated.

Concept Introduction:

Refer to part (a).

Answer to Problem 12.31P

Volume of $\text{0}{\text{.300 M H}}_2{\text{SO}}_{\text{4}}$ solution needed to neutralize $\text{25}\text{.0 }\mathrm{mL}$ solution of $\text{0}\text{.200 M NaOH}$ is $8.333\text{ mL}$.

Explanation of Solution

The dilution formula is given as follows:

  $n_1{M}_1{V}_1=n_2{M}_2{V}_2$        (1)

Substitute 2 for $n_1$, 1 for $n_2$, $\text{0}\text{.300 M}$ for $M_1$, $\text{0}\text{.200 M}$ for $M_2$, $\text{25}\text{.0 }\mathrm{mL}$ for $V_2$ in equation (1).

  $\left(2\right)\left(\text{0}\text{.300 M}\right)\left(V_1\right)=\left(1\right)\left(\text{0}\text{.200 M}\right)\left(\text{25}\text{.0 }\mathrm{mL}\right)$        (3)

Rearrange equation (3) to calculate the value of $V_1$.

  $\begin{array}{c}{V}_1=\frac{\left(\text{0}\text{.200 M}\right)\left(\text{25}\text{.0 }\mathrm{mL}\right)}{\left(2\right)\text{0}\text{.300 M}}\\ =8.333\text{ mL}\end{array}$