Respuesta :
1. This is an acid base neutralisation reaction ,
the balanced equation for the reaction is as follows
NaOH + HCl ---> NaCl + H₂O
molar ratio of NaOH to HCl is 1:1
number of moles are calculated by concentration x volume
therefore at neutralisation number of moles of HCl reacted is equal to number of NaOH moles
number of moles of NaOH reacted = 0.15 mol/dm³ x 67 x 10⁻³ dm³ = 10.05 x10⁻³ mol
the number of moles of HCl in 134 x 10⁻³ dm³ is 10.05 x 10⁻³ mol
to find the concentration of HCl , we need to find the number of moles in 1 dm³
in 134 x 10⁻³ dm³ there are - 10.05 x 10⁻³ mol
therefore in 1 dm³ there are- 10.05 x 10⁻³ mol / 134 x 10⁻³ dm³ = 0.075 mol/dm³
2. the acid base neutralisation reaction between KOH and H₂SO₄ is
2KOH + H₂SO₄ --> K₂SO₄ + 2H₂O
molar ratio of KOH to H₂SO₄ is 2:1
1 mol of H₂SO₄ reacts with 2 mol of KOH
the number of H₂SO₄ moles reacted - 0.050 mol/dm³ x 27.4 x 10⁻³ dm³ = 1.37 x 10⁻³ mol
number of KOH moles reacted is twice the number of H₂SO₄ moles reacted according to molar ratio
therefore number of HCl moles are - 1.37 x 10⁻³ x 2 = 2.74 x 10⁻³ mol
there are 2.74 x 10⁻³ mol of KOH in 357 x 10⁻³ dm³
therefore in 1 dm³ of KOH there are - 2.74 x 10⁻³ mol / 357 x 10⁻³ dm³
the concentration of KOH is - 0.00768 mol/dm³
even though its NaOH or KOH the concentration would be the same.
3. the balanced equation is
2NaOH + H₂SO₄ --> Na₂SO₄ + 2H₂O
molar ratio of NaOH to H₂SO₄ is 2:1
2 mol of NaOH reacts with 1 mol of H₂SO₄
number of NaOH moles reacted - 0.5 mol/dm³ x 55 x 10⁻³ dm³ = 25.5 x 10⁻³ mol
number of H₂SO₄ moles reacted is - half the number of NaOH moles reacted
number of H₂SO₄ moles reacted - 25.5 x 10⁻³ mol / 2 = 12.75 x 10⁻³ mol
there are 12.75 x 10⁻³ mol in 130 x 10⁻³ dm³
then in 1 dm³ there are - 12.75 x 10⁻³ mol / 130 x 10⁻³ dm³ = 0.098 mol/dm³
concentration is 0.098 M
