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Brrunno [24]
2 years ago
11

The maximum solubility of Ag2CO3 in 0.02M Na2CO3 is (Ksp ofAg2003 is 8 x 10-12)​

Chemistry
1 answer:
Anna007 [38]2 years ago
3 0

The maximum solubility of Ag₂CO₃ = = 1 x 10⁻⁵

<h3>Further explanation</h3>

Given

0.02 M Na₂CO₃

Ksp of  Ag₂CO₃ is 8 x 10⁻¹²

Required

The solubility of Ag₂CO₃

Solution

Ag₂CO₃ ⇒ 2Ag⁺ + CO₃²⁻

s                  2s        s

s = solubility

Ksp Ag₂CO₃ = [Ag⁺]² [CO₃²⁻]

Ksp Ag₂CO₃ = (2s)².s

Ksp Ag₂CO₃ = 4s³

In 0.02 M Na₂CO₃ ⇒ 2Na⁺ + CO₃²⁻⇒ [ CO₃²⁻]=0.02, so Ksp Ag₂CO₃ :

8 x 10⁻¹² = [2s]² [0.02]

8 x 10⁻¹² = 4s² [0.02]

4s² = 8 x 10⁻¹² / 2 × 10⁻²

4s² = 4 x 10⁻¹⁰

s² = 1 x 10⁻¹⁰

s = √1 x 10⁻¹⁰

s = 1 x 10⁻⁵

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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}\\

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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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Since volume and temperature are constant, this means that pressure and <u>number of moles</u> are <u>directly </u>proportional. the sample with the largest <u>number of moles</u> will have the <u>high </u>pressure.

Since, the ideal gas equation is also called ideal gas law. So,  according to ideal gas equations,

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P ∝ nR

since, R is also constant. So, conclusion of the final equation is

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The number of moles and pressure of the sample is directly proportion. So, on increasing number of moles in the sample , pressure of the sample also increases.

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