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 $pH = 3.4$:

$$[H^+] = 10^{-3.4}$$ $$[H^+] \approx 3.98 \times 10^{-4} \text{ M}$$

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.

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:

$$1.7 \times 10^{-5} = \frac{(3.98 \times 10^{-4})^2}{C_0 - (3.98 \times 10^{-4})}$$

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 $\text{0.0097 M}$.