Question

A solution is prepared by dissolving 0.56 g of benzoic acid (C6H5CO2H, Ka ???? 6.4 ???? 10????5) in enough water to make 1.0 L of solution. Calculate [C6H5CO2H], [C6H5CO2????], [H????], [OH????], and the pH of this solution.

Answer 1

Answer:

[H⁺] = 6.083x10⁻⁴ M, [C₆H₅OO⁻] = 6.083x10⁻⁴ M, [C₆H₅OOH] = 3.98x10⁻³M, pH = 3.22

Explanation:

Data: we have 0.56 gr of benzoic acid, disolved in 1Lt of water. Kₐ = 6.4x10⁻⁵

M (molar mass) of BA (Benzoic Acid) = 122 g/mol

Then, the inicial concentration is 0.56/122 = 4.59x10⁻³ M

We should consider the equation once it reaches the equilibrium:

C₆H₅COOH ⇄ C₆H₅COO⁻ + H⁺

  C - x                      x              x

And, for the Kₐ:

Kₐ = [H⁺][C₆H₅COO⁻]/[C₆H₅COOH] = x²/(C-x) , where C = 4.59x10⁻³

Then: x² + Kₐx - KₐC = 0

x² + 6.4x10⁻⁵ - 2.9x10⁻⁷ = 0

Resolving this cuadratic equation (remember to use Baskara equation), we obtain:

x = 6.083x10⁻⁴ M

Then: [H⁺] = [C₆H₅COO⁻] = 6.083x10⁻⁴ M

[C₆H₅COOH] = C - x = 3.98x10⁻³ M

pH = -Log [H⁺] = 3.22

Answer 2

Answer:

The values of

$\left[\mathrm{H}^{+}\right]=\left[\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2}^{-}\right]=5.1 \times 10^{-4} \mathrm{M}$

$\left[\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2} \mathrm{H}\right]=4.1 \times 10^{-3} \mathrm{M}$

$\left[\mathrm{OH}^{-}\right]=1.9 \times 10^{-11} \mathrm{M}$

$\mathrm{pH}=3.29$

Explanation:

Dissolving benzoic acid $=0.56 \ g$

$\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2} \mathrm{H}, K_{\mathrm{a}}=6.4 \times 10^{-5}\right)$

Water in solution $=1.0\ L$

Step 1:

Mass of benzoic acid dissolved in water $(\mathrm{m})=0.56$

Final volume $(V)=1 \ L$

The benzoic acid dissociation constant: $\mathrm{K}_{a}=6.4 \times 10^{-5}$

The number of moles of benzoic acid: $\mathrm{n}=\frac{\mathrm{m}}{\mathrm{M}_{m}}$

Molarity of benzoic acid: $\mathrm{M}=\frac{\mathrm{n}}{\mathrm{V}}$

Step 2:

Calculate the molarity (M) of benzoic acid.

For this, we must first calculate the molar mass $\left(\mathrm{M}_{m}\right)$ and the number of moles (n) of benzoic acid:

$\mathrm{M}_{m}=7 \times \mathrm{M}_{m}(\mathrm{C})+6 \times \mathrm{M}_{m}(\mathrm{H})+2 \times \mathrm{M}_{m}(\mathrm{O})$$\mathrm{M}_{m}=7 \times 12.011 \mathrm{~g} / \mathrm{mol}+6 \times 1.008 \mathrm{~g} / \mathrm{mol}+2 \times 15.999 \mathrm{~g} / \mathrm{mol}=122.123 \mathrm{~g} / \mathrm{mol}$

$\mathrm{n}=\frac{0.56 \mathrm{~g}}{122.123 \mathrm{~g} / \mathrm{mol}}=4.59 \times 10^{-3} \mathrm{~mol}$

$\mathrm{M}=\frac{4.59 \times 10^{-3} \mathrm{~mol}}{1 \mathrm{~L}}=4.59 \times 10^{-3} \mathrm{M}$

Step 3:

Dissociation equations and the dissociation constants:

$\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2} \mathrm{H}_{(a q)} \Leftrightarrow \mathrm{H}_{(a q)}^{+}+\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2}^{-}(a q) \quad \mathrm{K}_{a}=6.4 \times 10^{-5}$

$\mathrm{H}_{2} \mathrm{O}_{(l)} \Leftrightarrow \mathrm{H}_{(a q)}^{+}+\mathrm{OH}_{(a q)}^{-} \quad \mathrm{K}_{w}=1.0 \times 10^{-14}$

Step 4:

$\mathrm{K}_{a}=\frac{\left[\mathrm{H}^{+}\right]\left[\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2}^{-}\right]}{\left[\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2} \mathrm{H}\right]}=6.4 \times 10^{-5}$

The approximation for$\left[\mathrm{H}^{+}\right]_{0}$ is 0. (because of autonization)

$\left[\mathrm{H}^{+}\right]_{0}=10^{-7} \mathrm{M} \approx 0$

Step 5:

Symbol x represents the change in the concentrations.

$\left[\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2} \mathrm{H}\right]=\left[\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2} \mathrm{H}\right]_{0}-\mathrm{x}=0.00459-\mathrm{x}$

$\left[\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2}^{-}\right]=\left[\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2}^{-}\right]_{0}+\mathrm{x}=0+\mathrm{x}=\mathrm{x}$

$\left[\mathrm{H}^{+}\right]=\left[\mathrm{H}^{+}\right]_{0}+x=0+x=x$

Step 6:

Substituting x into the equilibrium expression:

$\frac{(x)(x)}{0.00459-x}=6.4 \times 10^{-5}$

We must determine if the $\left[\mathrm{H}^{+}\right]$ is much smaller than $0.00459$ to decide whether the following approximation is valid or not:

$0.00459-x \approx 0.00459$

$\frac{(x)(x)}{0.00459}=6.4 \times 10^{-5}$

$x^{2}=2.4 \times 10^{-7}$

$\mathrm{x}=4.9 \times 10^{-4}=\left[\mathrm{H}^{+}\right]$

$\frac{\mathrm{x}}{0.00459} \times 100 \%=10.7 \%>5 \%$

The approximation is not valid since more than $5 \%$ of acid is dissociated.

Step 7:

Calculating the concentrations of the species and the $\mathrm{pH}$ :

$\frac{(x)(x)}{0.00459-x}=6.4 \times 10^{-5}$

$x^{2}+6.4 \times 10^{-5} \times-2.9 \times 10^{-7}=0$

$a x^{2}+b x+c=0$

Use the quadratic formula to calculate .

$x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$

$\mathrm{x}=5.1 \times 10^{-4}$

$\mathrm{M}=\left[\mathrm{H}^{+}\right]=\left[\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2}^{-}\right]$

$\left[\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2} \mathrm{H}\right]=0.00459-0.00051=4.1 \times 10^{-3} \mathrm{M}$

$\left[\mathrm{OH}^{-}\right]=\frac{\mathrm{K}_{w}}{\left[\mathrm{H}^{+}\right]}=\frac{1.0 \times 10^{-14}}{5.1 \times 10^{-4}}=1.9 \times 10^{-11} \mathrm{M}$

$\mathrm{pH}=-\log \left[\mathrm{H}^{+}\right]=-\log \left(5.1 \times 10^{-4}\right)=3.29$

Thus we can say the results are:

$\left[\mathrm{H}^{+}\right]=\left[\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2}^{-}\right]=5.1 \times 10^{-4} \mathrm{M}$

$\left[\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2} \mathrm{H}\right]=4.1 \times 10^{-3} \mathrm{M}$

$\left[\mathrm{OH}^{-}\right]=1.9 \times 10^{-11} \mathrm{M}$

$\mathrm{pH}=3.29$

To learn more about pH of the solution, refer:

• https://brainly.com/question/21289565