Questions: Fill in the Blank 0.5 points Consider a 0.00550 M weak acid solution with a percent ionization of 8.20%. Round each answer to TWO places past the decimal in scientific notation and for pH. Part A: What is the concentration of H+ (in M) at equilibrium? Part B: What is the Ka of this weak acid? Hint: Set up an ICE table, determine the value of 'x', then plug that 'x' value back into the equilibrium expressions in Ka. Part C: What is the pH of this solution?

Solution Steps

The percent ionization of the weak acid is given as \(8.20\%\). This means that \(8.20\%\) of the initial concentration of the acid ionizes to form \([\mathrm{H}^+]\).

The initial concentration of the weak acid is \(0.00550 \, \text{M}\).

The concentration of \([\mathrm{H}^+]\) at equilibrium can be calculated as follows:

\[ \left[\mathrm{H}^+\right] = \frac{8.20}{100} \times 0.00550 \, \text{M} = 0.000451 \, \text{M} \]

Expressing this in scientific notation:

\[ \left[\mathrm{H}^+\right] = 4.51 \times 10^{-4} \, \text{M} \]

To find the \(K_a\), we use the expression for the equilibrium constant for a weak acid:

\[ K_a = \frac{[\mathrm{H}^+][\mathrm{A}^-]}{[\mathrm{HA}]} \]

Assuming the degree of ionization is small, the concentration of the acid at equilibrium is approximately the initial concentration minus the concentration of \([\mathrm{H}^+]\):

\[ [\mathrm{HA}] \approx 0.00550 - 0.000451 = 0.005049 \, \text{M} \]

Since \([\mathrm{H}^+] = [\mathrm{A}^-]\), we have:

\[ K_a = \frac{(0.000451)^2}{0.005049} = 4.03 \times 10^{-5} \]

The pH is calculated using the concentration of \([\mathrm{H}^+]\):

\[ \text{pH} = -\log_{10}(4.51 \times 10^{-4}) = 3.35 \]

Final Answer