Chemistry, 02.08.2019 14:30 jaay88

Ejercicios sobre análisis gravimétrico y análisis volumétrico. 3. calcule el ph de una solución que tiene una concentración de iones h3o+ de 3.6x10-4 m 4. calcule el volumen de h2so4 0.2m que se requieren para titular 23 ml de naoh 0.1m 5. calcule los mg de azul de metileno que se necesitan para preparar una solución de 0.011 m en 800 ml y con ese valor, indique el volumen que se requiere para hacer una dilución de 0.001 m y 0.00075 m. 6. determine el peso equivalente de una solución de kh2po4 0.02 n que se preparó en 0.125l ayuda con procedimiento y !

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Answer from: leah7378

1) Answer is: pH = 3,44.
c(H₃O⁺) = 3,6·10⁻⁴ M = 3,6·10⁻⁴ mol/L = 0,00036 mol/L.
pH = -logc(H₃O⁺).
pH = -log(0,00036 mol/L).
pH = 3,44.
When pH is less than seven (pH<), solution is acidic (like this example).
When is equal seven (pH = 7), solution is neutral.
When pH is greater than seven (pH > 7), solution is basic.

2) Answer is: volume of H₂SO₄ is 5,75 mL.
Chemical reaction: H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O.
c(H₂SO₄) = 0,2 M = 0,2 mol/L.
V(NaOH) = 23 mL = 0,023 L.
c(NaOH) = 0,1 M = 0,1 mol/L.
n(NaOH) = c(NaOH) · V(NaOH).
n(NaOH) = 0,1 mol/L · 0,023 L.
n(NaOH) = 0,0023 mol.
From chemical reaction: n(H₂SO₄) : n(NaOH) = 1 : 2.
n(H₂SO₄) = 0,00115mol.
V(H₂SO₄) = n(H₂SO₄) ÷ c(H₂SO₄).
V(H₂SO₄) = 0,00115 mol ÷ 0,2 mol/L.
V(H₂SO₄) = 0,00575 L = 5,75 mL.

3) c₁(solution) = 0,011 M = 0,011 mol/L.
V₁(solution) = 800 mL = 0,8 L.
M(methylene blue - C₁₆H₁₈ClN₃S) = 319,85 g/mol.
n₁ = c₁ · V₁.
n₁ = 0,011 mol/L · 0,8 L.
n₁ = 0,0088 mol.
m(C₁₆H₁₈ClN₃S) = 0,0088 mol · 319,85 g/mol.
m(C₁₆H₁₈ClN₃S) = 2,814 g.
m(C₁₆H₁₈ClN₃S) = 2,814 g · 1000mg/g = 2814 mg.
V = n ÷ c
V₂ = 0,0088 mol ÷ 0,001 mol/L = 8,8 L = 8800 mL.
V₃ = 0,0088 mol ÷ 0,00075 mol/L = 11,73 L = 11730 mL.

4) The normality or the equivalent concentration:
Cn(KH₂PO₄) = 0,02 N = 0,02 eq/L (equivalent per liter).
V(KH₂PO₄) = 0,125 L.
number of equivalents of solute = Cn(KH₂PO₄) · V(KH₂PO₄).
number of equivalents of solute = 0,02 eq/L · 0,125 L.
number of equivalents of solute = 0,0025 eq.
equivalent weight = M(KH₂PO₄) ÷ number of equivalents per mole of solute.
equivalent weight = 136,1 g/mol ÷ 1 eq/mol.
equivalent weight = 136,1 g/eq.

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