Acid-base reactions  (Page 4/5)

NaOH: V = 25 cm ${}^3$

HCl: V = 15 cm ${}^3$ ; C = 0.2 M

The equation is already balanced.

Therefore, n(HCl) = M $\times$ V (make sure that all the units are correct!)

M = 0.2mol.dm ${}^{-3}$

V = 15cm ${}^3$ = 0.015dm ${}^3$

Therefore

There are 0.003 moles of HCl that react

Look at the equation for the reaction. For every mole of HCl there is one mole of NaOH that is involved in the reaction. Therefore, if 0.003 moles of HCl react, we can conclude that the same quantity of NaOH is needed for the reaction. The number of moles of NaOH in the reaction is 0.003.

First convert the volume into dm ${}^3$ . V = 0.025 dm ${}^3$ . Then continue with the calculation.

The molarity of the NaOH solution is 0.12 mol.dm ${}^3$ or 0.12 M

${\mathrm{H}}_2{\mathrm{SO}}_4+2\mathrm{NaOH}\to {\mathrm{Na}}_2{\mathrm{SO}}_4+2{\mathrm{H}}_2\mathrm{O}$

M = n/V

V = 220 cm ${}^3$ = 0.22 dm ${}^3$

Therefore,

Remember that only 20 cm ${}^3$ of the sulfuric acid solution is used.

M = n/V, therefore n = M $\times$ V

According to the balanced chemical equation, the mole ratio of H ${}_2$ SO ${}_4$ to NaOH is 1:2. Therefore, the number of moles of NaOH that are neutralised is 0.0046 $\times$ 2 = 0.0092 mols.