Respuesta :
2NaOH + H₂SO₄ = Na₂SO₄ + 2H₂O
v(H₂SO₄)=0.25 L
c(H₂SO₄)=2.00 mol/L
v(NaOH)=2.00 L
n(H₂SO₄)=c(H₂SO₄)v(H₂SO₄)
n(NaOH)=2n(H₂SO₄)=2c(H₂SO₄)v(H₂SO₄)
c(NaOH)=n(NaOH)/v(NaOH)=2c(H₂SO₄)v(H₂SO₄)/v(NaOH)
c(NaOH)=2*2.00*0.25/2.00=0.5 mol/L
the concentration of solution of NaOH is 0.5 mol/L
v(H₂SO₄)=0.25 L
c(H₂SO₄)=2.00 mol/L
v(NaOH)=2.00 L
n(H₂SO₄)=c(H₂SO₄)v(H₂SO₄)
n(NaOH)=2n(H₂SO₄)=2c(H₂SO₄)v(H₂SO₄)
c(NaOH)=n(NaOH)/v(NaOH)=2c(H₂SO₄)v(H₂SO₄)/v(NaOH)
c(NaOH)=2*2.00*0.25/2.00=0.5 mol/L
the concentration of solution of NaOH is 0.5 mol/L
Answer: The molarity of the NaOH solution is 0.5 M.
Explanation:
Moles of [tex]H_2SO_4[/tex] in 0.25 L of 2.00 M solution:
[tex]Molarity=\frac{\text{Moles of }H_2SO_4}{\text{Volume in Liters of}H_2SO_4}[/tex]
[tex]2.00 M=\frac{\text{Moles of }H_2SO_4}{0.25 L}[/tex]
Moles of [tex]H_2SO_4[/tex] = 2.00 mol/L × 0.25 L = 0.5 moles
Moles of [tex]NaOH[/tex] in 2.00 L of an unknown Molarity :
[tex]H_2SO_4+2NaOH\rightarrow Na_2SO_4+2H_2O[/tex]
1 mole [tex]H_2SO_4[/tex] gives 2 moles of [tex]H^+[/tex] ions.
Then 0.5 mole [tex]H_2SO_4[/tex] will give: 0.5 mol × 2 = 1.0 mol
1.0 mol of [tex]H^+[/tex] will neutralize the 1.0 mol of [tex]OH^-[/tex] ions.
Moles of [tex]H^+[/tex] = Moles of [tex]OH^-[/tex] = 1.0 moles
1 mol of [tex]OH^-[/tex]are produced by 1 mol of NaOH
The 1 mole of [tex]OH^-[/tex]will be produced from = 1 mol of NaOH
[tex]Molarity=\frac{\text{Moles of }NaOH}{\text{Volume in Liters of NaOH}}[/tex]
[tex]M=\frac{1 mol}{2.00L}=0.5 mol/L=0.5M[/tex]
The molarity of the NaOH solution is 0.5 M.
