Q. 10.97 - An Introduction to General, Organic, and Biological Chemistry

Question

Calculate the volume, in milliliters, of a $0,150 \mathrm{MNaOH}$ solution that will complesely neutralize each of the following: (10.7)

a. $25.0 \mathrm{mL}$of a ().286 M HC7 solution

b. $10.0 \mathrm{mL}$of a $0.560 {\mathrm{MHSO}}_4$solution

c.  $5.00 \mathrm{mL}$ of a $0.618 \mathrm{M}$ HBr solution

Step-by-Step Solution

Answer

Therefore, $V_b=\left(\frac{M_s{V}_u}{n_a}\right)\left(\frac{n_b}{M_b}\right)$

$$\begin{gathered} =\left(\frac{(0.618\mathrm{M})(5.0 \mathrm{mL})}{\left(1\right)}\right)\left(\frac{\left(1\right)}{(0.150\mathrm{M})}\right) \\ =20.6 \mathrm{mL} \end{gathered}$$

$\therefore$ Volume of $\mathrm{NaOH}$ (of $0.150\mathrm{M}$ ) required$=20.6 \mathrm{mL}$

Step 1: Given (Part a)

That are given numbers are calculated by: (a)

Number of moles of acid that is reacted with one acid

Step 2: Explanation Part (a)

Formula: $\frac{M_a{V}_a}{n_a}=\frac{M_b{V}_b}{n_b}$

Where, $M_a,V_a,n_a$ are Morality, volume and number of moles of acid that is reacted with one mole of the base, respectively,

$M_b,V_b,n_b$are Morality, volume and number of moles of base that is reacted with one mole of acid, respectively,

(a) Given,$M_a=0.288 \mathrm{M}$

$V_a=25.0 \mathrm{mL}$

$$\begin{gathered} n_a=1 \\ {M}_b=0.1 \\ {V}_b=? \\ {n}_b=1 \end{gathered}$$

$M_b=0.150 \mathrm{M}$

Therefore, $V_b=\left(\frac{M_a{V}_a}{n_a}\right)\left(\frac{n_b}{M_b}\right)$

$$\begin{gathered} =\left(\frac{(0.288\mathrm{M})(25.0 \mathrm{mL})}{\left(1\right)}\right)\left(\frac{\left(1\right)}{(0.150\mathrm{M})}\right) \\ {V}_{\Delta }=48.0 \mathrm{mL} \end{gathered}$$

$\therefore$ Volume of$\mathrm{NaOH}$(of $0.150 \mathrm{M}$ ) required $=48.0 \mathrm{mL}$

Step 3: Given information Part (b)

That are given equations are: (b)

Step 4: Explanation Part (b)

(b) Given, $M_a=0.560 \mathrm{M}$

$$\begin{gathered} V_a=10.0 \mathrm{mL} \\ {n}_a=1 \\ {M}_b=0.150 \mathrm{M} \\ {V}_b=? \\ {n}_b=2 \end{gathered}$$

Therefore, $V_b=\left(\frac{M_a{V}_a}{n_a}\right)\left(\frac{n_b}{M_b}\right)$

$$\begin{gathered} =\left(\frac{(0.560\mathrm{M})(10.0 \mathrm{mL})}{\left(1\right)}\right)\left(\frac{\left(2\right)}{(0.150\mathrm{M})}\right) \\ =74.666 \mathrm{mL} \\ \approx 74.7 \mathrm{mL} \end{gathered}$$

$\therefore$ Volume of $\mathrm{NaOH}$ (of$0.150\mathrm{M}$ ) required$=74.7 \mathrm{mL}$

Step 5: Given information Part (c)

That are given equations are : (c)

Step 6: Explanation Part (c)

(c) Given,$M_a=0.618 \mathrm{M}$

$$\begin{gathered} V_a=5.0 \mathrm{mL} \\ {n}_a=1 \\ {M}_b=0.150 \mathrm{M} \\ {V}_b=? \\ {n}_b=1 \end{gathered}$$

Therefore, $V_b=\left(\frac{M_e{V}_a}{n_a}\right)\left(\frac{n_b}{M_b}\right)$

$$\begin{gathered} =\left(\frac{(0.618\mathrm{M})(5.0 \mathrm{mL})}{\left(1\right)}\right)\left(\frac{\left(1\right)}{(0.150\mathrm{M})}\right) \\ =20.6 \mathrm{mL} \end{gathered}$$

$\therefore$ Volume of $\mathrm{NaOH}$ (of $0.150 \mathrm{M}$ ) required$=20.6 \mathrm{mL}$