Question

Ejercicios sobre análisis gravimétrico y análisis volumétrico.

Answer 1

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.