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Solve for [tex]\( x \)[/tex]:
[tex]\[ 3x = 6x - 2 \][/tex]


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What is the equilibrium constant, [tex]\( K_{\text{eq}} \)[/tex], for the reaction:
[tex]\[ 2 \text{NOCl}(g) \rightarrow 2 \text{NO}(g) + \text{Cl}_2(g) \][/tex]
given the equilibrium concentrations:
[tex]\[ [\text{NO}] = 0.02\, M \][/tex]
[tex]\[ [\text{Cl}_2] = 0.01\, M \][/tex]
[tex]\[ [\text{NOCl}] = 0.5\, M \][/tex]

B. What is the pH of a weak acid solution with [tex]\([ \text{H}^+ ] = 2.1 \times 10^{-6}\, M\)[/tex]? (4 points)

Sagot :

GinnyAnsweridn
Sure! Let's solve the given two parts systematically:

### Part A: Calculate the Equilibrium Constant [tex]\( K_{\text{eq}} \)[/tex]

For the reaction:
[tex]\[ 2 \text{NOCl} (g) \rightleftharpoons 2 \text{NO} (g) + \text{Cl}_2 (g) \][/tex]

The expression for the equilibrium constant [tex]\( K_{\text{eq}} \)[/tex] is given by:
[tex]\[ K_{\text{eq}} = \frac{[\text{NO}]^2 [\text{Cl}_2]}{[\text{NOCl}]^2} \][/tex]

We're given the following equilibrium concentrations:
- [tex]\([ \text{NO} ] = 0.02 \, \text{M}\)[/tex]
- [tex]\([ \text{Cl}_2] = 0.01 \, \text{M}\)[/tex]
- [tex]\([ \text{NOCl} ] = 0.5 \, \text{M}\)[/tex]

Plugging these values into the equation:

[tex]\[ K_{\text{eq}} = \frac{(0.02)^2 \times 0.01}{(0.5)^2} \][/tex]

Calculating the numerator:
[tex]\[ (0.02)^2 = 0.0004 \][/tex]
[tex]\[ 0.0004 \times 0.01 = 0.000004 \][/tex]

Calculating the denominator:
[tex]\[ (0.5)^2 = 0.25 \][/tex]

So,
[tex]\[ K_{\text{eq}} = \frac{0.000004}{0.25} = 0.000016 \][/tex]
[tex]\[ K_{\text{eq}} = 1.6 \times 10^{-5} \][/tex]

Thus, the value of the equilibrium constant [tex]\( K_{\text{eq}} \)[/tex] is [tex]\( 1.6 \times 10^{-5} \)[/tex].

### Part B: Calculate the pH of a Weak Acid Solution

We're given the concentration of hydrogen ions [tex]\( [\text{H}^+] \)[/tex] as:
[tex]\[ [\text{H}^+] = 2.1 \times 10^{-6} \, \text{M} \][/tex]

The pH of a solution is calculated using the formula:
[tex]\[ \text{pH} = -\log_{10}([\text{H}^+]) \][/tex]

Substituting the given [tex]\( [\text{H}^+] \)[/tex]:

[tex]\[ \text{pH} = -\log_{10}(2.1 \times 10^{-6}) \][/tex]

Calculating the logarithm first:
[tex]\[ \log_{10}(2.1 \times 10^{-6}) = \log_{10}(2.1) + \log_{10}(10^{-6}) \][/tex]
[tex]\[ \log_{10}(2.1) \approx 0.322 \][/tex]
[tex]\[ \log_{10}(10^{-6}) = -6 \][/tex]

Adding these values:
[tex]\[ \log_{10}(2.1 \times 10^{-6}) = 0.322 - 6 = -5.678 \][/tex]

So, the pH is:
[tex]\[ \text{pH} = -(-5.678) = 5.678 \][/tex]

Therefore, the pH of the weak acid solution is approximately 5.68 (more precisely 5.6777807052660805).

### Summary
- The equilibrium constant [tex]\( K_{\text{eq}} \)[/tex] is [tex]\( 1.6 \times 10^{-5} \)[/tex].
- The pH of the weak acid solution is approximately 5.68.
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