Answer:
A. pH using molar concentrations = 2.56
B. pH using activities = 2.46
C. pH of mixture = 2.56
Explanation:
A. pH using molar concentrations
ClCH₂COOH + H₂O ⇌ ClCH₂COO⁻ + H₃O⁺
HA + H₂O ⇌ A⁻ + H₃O⁺
We have a solution of 0.08 mol HA and 0.04 mol A⁻
We can use the Henderson-Hasselbalch equation to calculate the pH.
![\begin{array}{rcl}\text{pH} & = & \text{pK}_{\text{a}} + \log \left(\dfrac{[\text{A}^{-}]}{\text{[HA]}}\right )\\\\& = & 2.865 +\log \left(\dfrac{0.04}{0.08}\right )\\\\& = & 2.865 + \log0.50 \\& = &2.865 - 0.30 \\& = & \mathbf{2.56}\\\end{array}](https://tex.z-dn.net/?f=%5Cbegin%7Barray%7D%7Brcl%7D%5Ctext%7BpH%7D%20%26%20%3D%20%26%20%5Ctext%7BpK%7D_%7B%5Ctext%7Ba%7D%7D%20%2B%20%5Clog%20%5Cleft%28%5Cdfrac%7B%5B%5Ctext%7BA%7D%5E%7B-%7D%5D%7D%7B%5Ctext%7B%5BHA%5D%7D%7D%5Cright%20%29%5C%5C%5C%5C%26%20%3D%20%26%202.865%20%2B%5Clog%20%5Cleft%28%5Cdfrac%7B0.04%7D%7B0.08%7D%5Cright%20%29%5C%5C%5C%5C%26%20%3D%20%26%202.865%20%2B%20%5Clog0.50%20%5C%5C%26%20%3D%20%262.865%20-%200.30%20%5C%5C%26%20%3D%20%26%20%5Cmathbf%7B2.56%7D%5C%5C%5Cend%7Barray%7D)
B. pH using activities
(i) Calculate [H⁺]
pH = -log[H⁺]
![\text{[H$^{+}$]} = 10^{-\text{pH}} \text{ mol/L} = 10^{-2.56}\text{ mol/L} = 2.73 \times 10^{-3}\text{ mol/L}](https://tex.z-dn.net/?f=%5Ctext%7B%5BH%24%5E%7B%2B%7D%24%5D%7D%20%20%3D%2010%5E%7B-%5Ctext%7BpH%7D%7D%20%5Ctext%7B%20mol%2FL%7D%20%3D%2010%5E%7B-2.56%7D%5Ctext%7B%20mol%2FL%7D%20%3D%202.73%20%20%5Ctimes%2010%5E%7B-3%7D%5Ctext%7B%20mol%2FL%7D)
(ii) Calculate the ionic strength of the solution
We have a solution of 0.08 mol·L⁻¹ HA, 0.04 mol·L⁻¹ Na⁺, 0.04 mol·L⁻¹ A⁻, and 0.00273 mol·L⁻¹ H⁺.
The formula for ionic strength is
![I = \dfrac{1}{2} \sum_{i} {c_{i}z_{i}^{2}}\\\\I = \dfrac{1}{2}\left [0.04\times (+1)^{2} + 0.04\times(-1)^{2} + 0.00273\times(+1)^{2}\right]\\\\= \dfrac{1}{2} (0.04 + 0.04 + 0.00273) = \dfrac{1}{2} \times 0.08273 = 0.041](https://tex.z-dn.net/?f=I%20%3D%20%5Cdfrac%7B1%7D%7B2%7D%20%5Csum_%7Bi%7D%20%7Bc_%7Bi%7Dz_%7Bi%7D%5E%7B2%7D%7D%5C%5C%5C%5CI%20%3D%20%5Cdfrac%7B1%7D%7B2%7D%5Cleft%20%5B0.04%5Ctimes%20%28%2B1%29%5E%7B2%7D%20%2B%200.04%5Ctimes%28-1%29%5E%7B2%7D%20%2B%20%200.00273%5Ctimes%28%2B1%29%5E%7B2%7D%5Cright%5D%5C%5C%5C%5C%3D%20%20%5Cdfrac%7B1%7D%7B2%7D%20%280.04%20%2B%200.04%20%2B%200.00273%29%20%3D%20%5Cdfrac%7B1%7D%7B2%7D%20%5Ctimes%200.08273%20%3D%200.041)
(iii) Calculate the activity coefficients
![\ln \gamma = -0.510z^{2}\sqrt{I} = -0.510(-1)^{2}\sqrt{0.041} = -0.510\times 0.20 = -0.10\\\gamma = 10^{-0.10} = 0.79](https://tex.z-dn.net/?f=%5Cln%20%5Cgamma%20%3D%20-0.510z%5E%7B2%7D%5Csqrt%7BI%7D%20%3D%20-0.510%28-1%29%5E%7B2%7D%5Csqrt%7B0.041%7D%20%3D%20-0.510%5Ctimes%200.20%20%3D%20-0.10%5C%5C%5Cgamma%20%3D%2010%5E%7B-0.10%7D%20%3D%200.79)
(iv) Calculate the initial activity of A⁻
a = γc = 0.79 × 0.04= 0.032
(v) Calculate the pH
![\begin{array}{rcl}\text{pH} & = & \text{pK}_{\text{a}} + \log \left(\dfrac{a_{\text{A}^{-}}}{a_{\text{[HA]}}}\right )\\\\& = & 2.865 +\log \left(\dfrac{0.032}{0.08}\right )\\\\& = & 2.865 + \log0.40 \\& = & 2.865 -0.40\\& = & \mathbf{2.46}\\\end{array}\\](https://tex.z-dn.net/?f=%5Cbegin%7Barray%7D%7Brcl%7D%5Ctext%7BpH%7D%20%26%20%3D%20%26%20%5Ctext%7BpK%7D_%7B%5Ctext%7Ba%7D%7D%20%2B%20%5Clog%20%5Cleft%28%5Cdfrac%7Ba_%7B%5Ctext%7BA%7D%5E%7B-%7D%7D%7D%7Ba_%7B%5Ctext%7B%5BHA%5D%7D%7D%7D%5Cright%20%29%5C%5C%5C%5C%26%20%3D%20%26%202.865%20%2B%5Clog%20%5Cleft%28%5Cdfrac%7B0.032%7D%7B0.08%7D%5Cright%20%29%5C%5C%5C%5C%26%20%3D%20%26%202.865%20%2B%20%5Clog0.40%20%5C%5C%26%20%3D%20%26%202.865%20-0.40%5C%5C%26%20%3D%20%26%20%5Cmathbf%7B2.46%7D%5C%5C%5Cend%7Barray%7D%5C%5C)
C. Calculate the pH of the mixture
The mixture initially contains 0.08 mol HA, 0.04 mol Na⁺, 0.04 mol A⁻, 0.05 mol HNO₃, and 0.06 mol NaOH.
The HNO₃ will react with the NaOH to form 0.05 mol Na⁺ and 0.05 mol NO₃⁻.
The excess NaOH will react with 0.01 mol HA to form 0.01 mol Na⁺ and 0.01 mol A⁻.
The final solution will contain 0.07 mol HA, 0.10 mol Na⁺, 0.05 mol A⁻, and 0.05 mol NO₃⁻.
(i) Calculate the ionic strength
![I = \dfrac{1}{2}\left [0.10\times (+1)^{2} + 0.05 \times(-1)^{2} + 0.05\times(-1)^{2}\right]\\\\= \dfrac{1}{2} (0.10 + 0.05 + 0.05) = \dfrac{1}{2} \times 0.20 = 0.10](https://tex.z-dn.net/?f=I%20%3D%20%5Cdfrac%7B1%7D%7B2%7D%5Cleft%20%5B0.10%5Ctimes%20%28%2B1%29%5E%7B2%7D%20%2B%200.05%20%5Ctimes%28-1%29%5E%7B2%7D%20%2B%20%200.05%5Ctimes%28-1%29%5E%7B2%7D%5Cright%5D%5C%5C%5C%5C%3D%20%20%5Cdfrac%7B1%7D%7B2%7D%20%280.10%20%2B%200.05%20%2B%200.05%29%20%3D%20%5Cdfrac%7B1%7D%7B2%7D%20%5Ctimes%200.20%20%3D%200.10)
(ii) Calculate the activity coefficients
![\ln \gamma = -0.510z^{2}\sqrt{I} = -0.510(-1)^{2}\sqrt{0.10} = -0.510\times 0.32 = -0.16\\\gamma = 10^{-0.16} = 0.69](https://tex.z-dn.net/?f=%5Cln%20%5Cgamma%20%3D%20-0.510z%5E%7B2%7D%5Csqrt%7BI%7D%20%3D%20-0.510%28-1%29%5E%7B2%7D%5Csqrt%7B0.10%7D%20%3D%20-0.510%5Ctimes%200.32%20%3D%20-0.16%5C%5C%5Cgamma%20%3D%2010%5E%7B-0.16%7D%20%3D%200.69)
(iii) Calculate the initial activity of A⁻:
a = γc = 0.69 × 0.05= 0.034
(iv) Calculate the pH
![\text{pH} = 2.865 + \log \left(\dfrac{0.034}{0.07}\right ) = 2.865 + \log 0.49 = 2.865 - 0.31 = \mathbf{2.56}](https://tex.z-dn.net/?f=%5Ctext%7BpH%7D%20%3D%202.865%20%2B%20%5Clog%20%5Cleft%28%5Cdfrac%7B0.034%7D%7B0.07%7D%5Cright%20%29%20%3D%202.865%20%2B%20%5Clog%200.49%20%3D%202.865%20-%200.31%20%3D%20%5Cmathbf%7B2.56%7D)