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Anit [1.1K]
2 years ago
7

We mix 0.08 moles of chloroacetic acid (ClCH2COOH) and 0.04 moles of

Chemistry
1 answer:
Arte-miy333 [17]2 years ago
3 0

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}

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}

(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

(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

(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}\\

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

(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

(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}

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Answer:

  • <em>The solution that has the highest concentration of hydroxide ions is </em><u>d. pH = 12.59.</u>

Explanation:

You can solve this question using just some chemical facts:

  1. pH is a measure of acidity or alkalinity: the higher the pH the lower the acidity and the higher the alkalinity.
  2. The higher the concentration of hydroxide ions the lower the acidity or the higher the alkalinity of the solution, this is the higher the pH.

Hence, since you are asked to state the solution with the highest concentration of hydroxide ions, you just pick the highest pH. This is the option d, pH = 12.59.

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Solution    pH        pOH                            [OH⁻]

a.               3.21       14 - 3.21 = 10.79        antilogarithm of 10.79 = 1.6 × 10⁻¹¹

b.               7.00      14 - 7.00 = 7.00        antilogarithm of 7.00 = 10⁻⁷

c.                7.93      14 - 7.93 = 6.07        antilogarithm of 6.07 = 8.5 × 10⁻⁷

d.               12.59     14 - 12.59 = 1.41        antilogarithm of 1.41 = 0.039

e.               9.82      14 - 9.82 = 4.18        antilogarithm of 4.18 = 6.6 × 10⁻⁵

From which you see that the highest concentration of hydroxide ions is for pH = 12.59.

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