Question

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

A. 250.0 mL of pure wateree.

B. 250.0 mL of a buffer solution that is 0.195 M in HCHO2 and 0.275 M in KCHO

C. 250.0 mL of a buffer solution that is 0.255 M in CH3CH2NH2 and 0.235 M in CH3CH2NH3Cl

Answer 1

To calculate the final pH after the addition of NaOH, consider the initial pH, buffer components' concentrations, and acid-base reactions. Pure water's final pH remains close to 7. Buffer solutions resist large pH changes, so the final pH of a buffer depends on the equilibrium between the acid and its conjugate base. In a buffer solution of a weak base and its conjugate acid, NaOH's addition leads to a slight decrease in pH.

Step-by-step explanation:

In order to calculate the final pH after the addition of NaOH, we need to consider the initial pH, the concentration of the buffer components, and the reaction between the acid-base pairs. Let's go through each situation:

A. Pure Water:

Since pure water is neutral, the initial pH is 7. After the addition of NaOH, it dissociates into Na+ and OH- ions. The OH- ions react with H+ ions in water to form water, which does not affect the pH significantly. Therefore, the final pH remains close to 7.

B. Buffer solution (HCHO2/KCHO):

First, we need to determine which species will react with NaOH. In this case, HCHO2 will react with NaOH to form water and HCO2-. The KCHO does not react with NaOH. The initial pH of the buffer solution will depend on the equilibrium between HCHO2 and HCO2-. After the addition of NaOH, the concentration of HCHO2 decreases and the concentration of HCO2- increases, resulting in a shift in the equilibrium. This will cause a slight increase in pH, but the buffer solution will resist large changes in pH.

C. Buffer solution (CH3CH2NH2/CH3CH2NH3Cl):

In this buffer solution, CH3CH2NH2 is the weak base and CH3CH2NH3Cl is its conjugate acid. The initial pH depends on the equilibrium between the weak base and its conjugate acid. After the addition of NaOH, the weak base reacts to form CH3CH2NH2OH, which does not significantly affect the pH. The equilibrium will shift to reduce the concentration of the weak base and increase the concentration of its conjugate acid, resulting in a slight decrease in pH.

Answer 2

A. 12.6

B. 4.05

C. 10.93

Step-by-step 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