Question

For each solution, calculate the initial and final pH after the addition of 0.010 mol of NaOH.

Answer 1

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