Question

The equation shows the reaction of sulfuric acid with sodium hydroxide:

H2SO4 + 2NaOH --> NaSO4 + 2H2O
25 cm^3 of sulfuric acid of concentration 0.2 mol/dm^3 reacted with exactly 10 cm^3 of sodium hydroxide. Calculate:
a) the number of moles of sulfuric acid present
b) the number of moles of sodium hydroxide reacting
c) the concentration of the sodium hydroxide in mol/dm^3

Answer 1

(a).

`{ \sf{1000 \: {cm}^(3) \: contains \: 0.2 \: moles \: of \: acid}} \\ { \sf{25 \: {cm}^(3) \: contains \: ( (25 * 0.2)/(1000) ) \: moles}} \\ \\ = { \sf{0.005 \: moles}}`

» 0.005 moles of sulphuric acid are present.

(b).

from the equation:

[ 1 mole of acid reacts with 2 moles of sodium hydroxide ]

`{ \sf{1 \: mole \: of \: acid \: reacts \: with \: 2 \: moles \: of \: hydroxide}} \\ { \sf{0.005 \: moles \: of \: acid \: react \: with \: (0.005 * 2) \: moles}} \\ { \sf{ = 0.01 \: moles}}`

» 0.01 moles of sodium hydroxide reacted.

(c).

`{ \sf{10 \: {cm}^(3) \: contains \: 0.01 \: moles \: of \: hydroxide }} \\ { \sf{1000 \: {cm}^( 3) \: contain \: ( (1000 * 0.01)/(10) ) \: moles}} \\ \\ { \sf{ = 1 \: mol \: {l}^( - 1) }}`

» concentration is 1 M