Question

For which of the following solutions, a minimum amount of $NaOH$ is required for complete neutrallsation?
[Assume complete dissociation of acid]

• A. $50\text{mL}{H}_{2}S{O}_4$ solution [basicity = 2]
  $(d_{\text{solution}}=2\text{g/mL})$, % w/w = 98
• B. $50\text{mL}$ $H_{3}P{O}_4$ solution [basicity =3]
  $(d_{\text{solution}}=2\text{g/mL})$, % w/w = 98
• C. $100\text{mL}HCl$ solution [basicity = 1]
  $(d_{\text{solution}}=1.5\text{g/mL})$, % w/w = 36.5
• D. $100\text{mL}HN{O}_3$ solution [basicity =1]
  $(d_{\text{solution}}=1.5\text{g/mL})$ , % w/w =36.5

Answer 1

The correct option is D $100\text{mL}HN{O}_3$ solution [basicity =1]

$(d_{\text{solution}}=1.5\text{g/mL})$ , % w/w =36.5
(a) $H_{2}S{O}_4+2NaOH\to N{a}_{2}S{O}_4+2H_{2}O$

98% w/w means 98 g of $H_{2}S{O}_4$ is present in 100 g of the solution.
$\text{density}=\frac{\text{mass of solution}}{\text{volume of solution}}$
$2=\frac{100}{V}$
$V=50\text{mL}$
So, the volume should be $50\text{mL}$ for $100g$ of the solution and also $50\text{mL}$ is given in the question.
So, moles of $H_{2}S{O}_4=\frac{98}{98}=1$
From the above balanced equation,
Moles of $NaOH=2\times$ moles of $H_{2}S{O}_4=2$

(b) $H_{3}P{O}_4+3NaOH\to N{a}_{3}P{O}_4+3H_{2}O$

98% w/w means 98 g of $H_{3}P{O}_4$ is present in 100 g of the solution.
$\text{density}=\frac{\text{mass of solution}}{\text{volume of solution}}$
$2=\frac{100}{V}$
$V=50\text{mL}$
So, the volume should be $50\text{mL}$ for $100g$ of the solution and also $50\text{mL}$ is given in the question.
So, moles of $H_{3}P{O}_4=\frac{98}{98}=1$
From the above balanced equation,
Moles of $NaOH=3\times$ moles of $H_{3}P{O}_4=3$

(c) $HCl+NaOH\to NaCl+H_{2}O$

36.5% w/w means 36.5 g of $HCl$ is present in 100 g of the solution.
$\text{density}=\frac{\text{mass of solution}}{\text{volume of solution}}$
$1.5=\frac{100}{V}$
$V=66.66\text{mL}$
So, the volume should be $66.66\text{mL}$ for $100g$ of the solution, but we have $100\text{mL}$ given in the question.
So, $66.66\text{mL}$ contain $36.5g$
$100\text{mL}$ will contain $=\frac{36.5\times 100}{66.66}=54.7g$
So, moles of $HCl=\frac{54.7}{36.5}=1.5$
From the above balanced equation,
Moles of $NaOH$ = moles of $HCl=1.5$

(d) $HN{O}_3+NaOH\to NaN{O}_3+H_{2}O$

36.5% w/w means 36.5 g of $HN{O}_3$ is present in 100 g of the solution.
$\text{density}=\frac{\text{mass of solution}}{\text{volume of solution}}$
$1.5=\frac{100}{V}$
$V=66.66\text{mL}$
So, the volume should be $66.66\text{mL}$ for $100g$ of the solution, but we have $100mL$ given in the question.
So, $66.66\text{mL}$ contain $36.5g$
$100\text{mL}$ will contain $=\frac{36.5\times 100}{66.66}=54.7g$
So, moles of $HN{O}_3=\frac{54.7}{63}=0.87$
From the above balanced equation,
Moles of NaOH = moles of $HN{O}_3=0.87$

Hence least amount is required in (d)