Question

2.42.10cm3 of a solution containing 2g of an acid (H2X) per dm3 neutralizes 28cm3 of solution containing 2.5g of potassium hydroxide per dm3. Calculate
i.The concentration of the acid in moles/dm3
ii.The molar mass of the acid H2X
iii.The value of X
iv.The percentage by mass of X

Answer 1

The balanced equation for the reaction of H₂X with potassium hydroxide is:

H₂X + 2KOH $\rightarrow$ K₂X + 2H₂O.

According to the stoichiometry coefficients, 1 mol of H₂X reacts with 2 mol of KOH:

$n(H_2X) = \frac{n(KOH)}{2}$ .

The molar concentration of KOH solution is:

$c = \frac{n}{V} = \frac{m}{MV} = \frac{2.5(g)}{56.11(g/mol)·1(dm^3)} = 0.0446$ mol/dm³.

The number of the moles of potassium hydroxide reacted is the product of the concentration and the volume of its solution:

$n(KOH) = cV$

$n(KOH)= 0.0446(mol/dm^3)·28·10^{-3} (dm^3)$

$n(KOH) = 1.25·10^{-3}$ mol.

Therefore, the number of the moles of the H₂X acid is:

$n(H_2X) = \frac{1.25·10^{-3}}{2} = 0.62·10^{-3}$ mol.

The molar concentration of the acid is the number of the moles divided by its volume:

$c(H_2X) = \frac{n}{V} = \frac{0.62·10^{-3}(mol)}{42.10·10^{-3}(dm^3)} = 0.0148$ mol/dm³.

The molar mass of H₂X can be calculated from its molar concentration and mass concentration (g/dm³):

$M = \frac{m}{V}·\frac{1}{c} = 2(g/dm^3)·\frac{1}{0.0148(mol/dm^3)} = 135$ g/mol.

The molar mass of X is then:

$M_X = M(H_2X) - 2M_H = 135-2 = 133$ g/mol.

Using the data given, X can't be assigned to any element in the periodic table.

The percentage by mass of X in the molecule H₂X is:

$\frac{M(X)}{M(H_2X)}·100\% = \frac{133}{135}·100\% =98.5\%$ .

Answer:

i.The concentration of the acid is 0.0148 mol/dm³

ii.The molar mass of the acid H₂X is 135 g/mol

iii.The data given is not consistent and doesn't permit to find the value of X

iv.The percentage by mass of X is 98.5%