Respuesta :
From the equation for molarity (number of moles of a substance dissolved in one liter of solution), we can get an equation for determining the number of moles:
c=n/V ⇒ n=c×V
Now, we can write required equations:
A n(NaOH)=c(NaOH)×V(NaOH)
B n(H₂SO₄)=c(H₂SO₄)×V(H₂SO₄)
From the reaction, we can conclude that NaOH and H₂SO₄ have following stoichiometric ratio:
2NaOH + H₂SO₄ ⇒ Na₂SO₄ + 2H₂0
n(NaOH) : n(H₂SO₄) = 2 : 1
So, we can write:
n(NaOH) = 2 × n(H₂SO₄)
And if we connect the above equations, we get the equation for solving c(H2SO4):
c(NaOH)×V(NaOH) = 2×c(H₂SO₄)×V(H₂SO₄)
c(H₂SO₄) = (c(NaOH)×V(NaOH)) / (2×V(H₂SO₄))
c=n/V ⇒ n=c×V
Now, we can write required equations:
A n(NaOH)=c(NaOH)×V(NaOH)
B n(H₂SO₄)=c(H₂SO₄)×V(H₂SO₄)
From the reaction, we can conclude that NaOH and H₂SO₄ have following stoichiometric ratio:
2NaOH + H₂SO₄ ⇒ Na₂SO₄ + 2H₂0
n(NaOH) : n(H₂SO₄) = 2 : 1
So, we can write:
n(NaOH) = 2 × n(H₂SO₄)
And if we connect the above equations, we get the equation for solving c(H2SO4):
c(NaOH)×V(NaOH) = 2×c(H₂SO₄)×V(H₂SO₄)
c(H₂SO₄) = (c(NaOH)×V(NaOH)) / (2×V(H₂SO₄))
The concentration of [H2SO4] is obtained from; [NaOH] × VNaOH/ 2× VH2SO4
The equation of the reaction is;
2NaOH(aq) + H2SO4(aq) ------> Na2SO4(aq) + 2H2O(l)
Number of moles of H2SO4(nH2SO4) = [H2SO4] × VH2SO4
Number of moles of NaOH(nNaOH) = [NaOH] × VNaOH
Where;
[H2SO4] = concentration of H2SO4
VH2SO4 = volume of H2SO4
[NaOH] = concentration of NaOH
VNaOH = volume of NaOH
From the balanced reaction equation;
2 moles of NaOH reacts with 1 mole of H2SO4
2nH2SO4 = nNaOH
2[H2SO4] × VH2SO4 = [NaOH] × VNaOH
[H2SO4] = [NaOH] × VNaOH/ 2× VH2SO4
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