Answer to Question #120843 in Chemistry for Paul

Question

1.30cm3 of hydrochloric acid solution was titrated against 41.20cm3 of 0.05M sodium hydroxide solution. Calculate
i.The concentration of the acid in moles/dm3
ii.The concentration of the acid in grammes per dm3

Answer 1

The reaction equation of the titration of hydrochloric acid against sodium hydroxide is:

HCl + NaOH $\rightarrow$ NaCl + H₂O.

According to the stoichiometric coefficients, 1 mol of hydrochloric acid reacts with 1 mol of sodium hydroxide. Therefore, the number of the moles of HCl in 30 cm³ of its solution equals the amount of NaOH in 41.20cm³ of its 0.05M solution:

$n(HCl) = n(NaOH)$ .

According to the definition of the molar concentration, the number of the moles $n$ contained in a volume of the solution $V$ with a molar concentration $c$ is:

$n = c·V$ .

Therefore, the molar concentration of hydrochloric acid solution is:

$c(HCl)·V(HCl) = c(NaOH)·V(NaOH)$

$c(HCl) = \frac{c(NaOH)·V(NaOH)}{V(HCl)}$

$c(HCl) = \frac{0.05·41.20}{30} = 0.07$ M, or 0.07 mol·dm^-3.

The concentration of the acid in grams per dm³ is the ratio of the mass of the solute to the volume of the solution. Making use of the relation between the number of the moles and the mass, you see that g/dm³ is the molar concentration times the molar mass of HCl (36.46 g/mol):

$c_m = \frac{m}{V} = \frac{nM}{V} = c·M$ .

$c_m = 0.07·36.46 = 2.5$ g/dm³.

Answer: i.The concentration of the acid in moles/dm³ is 0.07 mol/dm³;

ii.The concentration of the acid in grammes per dm3 is 2.5 g/dm³.