Here's how to calculate the initial concentration of the weak acid HA: Step 1: Calculate the equilibrium concentration of hydrogen ions, $[H^+]$, from the given pH. The formula for pH is: $$pH = -\log_{10}[H^+]$$ Rearranging to solve for $[H^+]$: $$[H^+] = 10^{-pH}$$ Given $pH = 3.4$: $$[H^+] = 10^{-3.4}$$ $$[H^+] \approx 3.98 \times 10^{-4} \text{ M}$$ Step 2: Write the dissociation equilibrium for the weak acid HA and set up an ICE table. The dissociation of a weak acid HA is: $$HA(aq) \rightleftharpoons H^+(aq) + A^-(aq)$$ Let $C_0$ be the initial concentration of HA. | | HA | $H^+$ | $A^-$ | | :---------- | :----------: | :----------: | :----------: | | Initial (I) | $C_0$ | $0$ | $0$ | | Change (C) | $-x$ | $+x$ | $+x$ | | Equilibrium (E) | $C_0 - x$ | $x$ | $x$ | From Step 1, we know that at equilibrium, $[H^+] = x = 3.98 \times 10^{-4}$ M. Therefore, $[A^-] = x = 3.98 \times 10^{-4}$ M. Step 3: Write the acid dissociation constant ($K_a$) expression and substitute the equilibrium concentrations. The $K_a$ expression is: $$K_a = \frac{[H^+][A^-]}{[HA]}$$ Substitute the equilibrium concentrations from the ICE table: $$K_a = \frac{(x)(x)}{C_0 - x} = \frac{x^2}{C_0 - x}$$ Given $K_a = 1.7 \times 10^{-5}$ and $x = 3.98 \times 10^{-4}$ M: $$1.7 \times 10^{-5} = \frac{(3.98 \times 10^{-4})^2}{C_0 - (3.98 \times 10^{-4})}$$ Step 4: Solve for the initial concentration, $C_0$. Rearrange the equation to solve for $C_0$: $$K_a (C_0 - x) = x^2$$ $$K_a C_0 - K_a x = x^2$$ $$K_a C_0 = x^2 + K_a x$$ $$C_0 = \frac{x^2 + K_a x}{K_a}$$ $$C_0 = \frac{x^2}{K_a} + x$$ Substitute the values: $$C_0 = \frac{(3.98 \times 10^{-4})^2}{1.7 \times 10^{-5}} + (3.98 \times 10^{-4})$$ $$C_0 = \frac{1.58404 \times 10^{-7}}{1.7 \times 10^{-5}} + 3.98 \times 10^{-4}$$ $$C_0 = 0.00931788 + 0.000398$$ $$C_0 = 0.00971588 \text{ M}$$ Rounding to two significant figures (based on the given pH and $K_a$ values): $$C_0 \approx 0.0097 \text{ M}$$ The initial concentration of the weak acid HA is $\boxed{\text{0.0097 M}}$.