91Ó°ÊÓ

That are given numbers are calculated by: (a)

Number of moles of acid that is reacted with one acid

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

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

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

That are given equations are: (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, $V_b=\left(\frac{M_a{V}_a}{n_a}\right)\left(\frac{n_b}{M_b}\right)$

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

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

That are given equations are : (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, $V_b=\left(\frac{M_e{V}_a}{n_a}\right)\left(\frac{n_b}{M_b}\right)$

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

$∴$Volume of $\mathrm{NaOH}$(of localid="1654586203294" $0.150\mathrm{M}$) required$=20.6\mathrm{mL}$