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 = <em>12.6</em>
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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]
<em>pH = 4.05</em>
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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]
<em>pH = 10.93</em>
