Respuesta :
Answer:
A. 12.6
B. 4.05
C. 10.93
Explanation:
A. 0.010mol of NaOH in 250.0mL gives a concentration of 0.04M NaOH = 0.04M OH⁻
pOH = -log [OH⁻] = 1.398
pH = 14-pOH = 12.6
B. The reaction of NaOH with HCHO₂ is:
NaOH + HCHO₂ → H₂O + CHO₂⁻ + Na⁺
Initial moles of CHO₂⁻ and HCHO₂ are:
CHO₂⁻ = 0.250L × (0.275mol/L) = 0.06875moles
HCHO₂ = 0.250L × (0.195mol/L) = 0.04875moles
After reaction:
CHO₂⁻ = 0.06875moles + 0.010mol = 0.07875mol
HCHO₂ = 0.04875moles - 0.010mol = 0.03875mol
Using H-H equation (pKa of this buffer: 3.74)
pH = 3.74 + log [CHO₂⁻] / [HCHO₂]
pH = 3.74 + log [0.07875mol] / [0.03875mol]
pH = 4.05
C. The reaction of NaOH with CH₃CH₂NH₃Cl is:
NaOH + CH₃CH₂NH₃Cl → H₂O + CH₃CH₂NH₂ + Cl⁻ + Na⁺
Initial moles of CH₃CH₂NH₃Cl and CH₃CH₂NH₂ are:
CH₃CH₂NH₃Cl = 0.250L × (0.235mol/L) = 0.05875moles
CH₃CH₂NH₂ = 0.250L × (0.255mol/L) = 0.06375moles
After reaction:
CH₃CH₂NH₃Cl = 0.05875moles - 0.010mol = 0.04875mol
CH₃CH₂NH₂ = 0.06375moles + 0.010mol = 0.07375mol
Using H-H equation (pKa of this buffer: 10.75)
pH = 10.75 + log [CH₃CH₂NH₂] / [CH₃CH₂NH₃Cl]
pH = 10.75 + log [0.07375mol] / [0.04875mol]
pH = 10.93
(A) The final pH for 250.0 mL of pure water is 12.6
(B) The final pH for the buffer solution is 4.05
(C) The final pH is 10.93
Calculating the pH of the solution:
(A) The concentration of the compounds in 0.010mol of NaOH in 250.0mL is :
0.010×250/1000 = 0.04M NaOH = 0.04M OH⁻
pH = -log [OH⁻] = 1.398
pH = 14-pOH = 12.6
(B) The equation of the reaction of NaOH with HCHO₂ is given by:
NaOH + HCHO₂ → H₂O + CHO₂⁻ + Na⁺
Initial amounts of CHO₂⁻ and HCHO₂ are:
CHO₂⁻ = 0.250L × (0.275mol/L) = 0.06875moles
HCHO₂ = 0.250L × (0.195mol/L) = 0.04875moles
The final amounts of the compounds is :
CHO₂⁻ = 0.06875moles + 0.010mol = 0.07875mol
HCHO₂ = 0.04875moles - 0.010mol = 0.03875mol
Using H-H equation (pKa of this buffer: 3.74)
pH = 3.74 + log [CHO₂⁻] / [HCHO₂]
pH = 3.74 + log [0.07875mol] / [0.03875mol]
pH = 4.05
(C) The equation of the reaction of NaOH with CH₃CH₂NH₃Cl is:
NaOH + CH₃CH₂NH₃Cl → H₂O + CH₃CH₂NH₂ + Cl⁻ + Na⁺
Initial moles of CH₃CH₂NH₃Cl and CH₃CH₂NH₂ are:
CH₃CH₂NH₃Cl = 0.250L × (0.235mol/L) = 0.05875moles
CH₃CH₂NH₂ = 0.250L × (0.255mol/L) = 0.06375moles
The final amounts of the compounds is :
CH₃CH₂NH₃Cl = 0.05875moles - 0.010mol = 0.04875mol
CH₃CH₂NH₂ = 0.06375moles + 0.010mol = 0.07375mol
Using the H-H equation (pKa of this buffer: 10.75) we get:
pH = 10.75 + log [CH₃CH₂NH₂] / [CH₃CH₂NH₃Cl]
pH = 10.75 + log [0.07375mol] / [0.04875mol]
pH = 10.93
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