First, we calculate the molarity of the $\text{NaOH}$ solution. Molar mass of $\text{NaOH} = 23 + 16 + 1 = 40 \text{ g/mol}$. Number of moles of $\text{NaOH} = \frac{\text{Given mass}}{\text{Molar mass}} = \frac{4 \text{ g}}{40 \text{ g/mol}} = 0.1 \text{ mol}$. Volume of solution $= 1000 \text{ ml} = 1 \text{ L}$. Molarity ($M$) $= \frac{\text{Moles of solute}}{\text{Volume of solution in L}} = \frac{0.1}{1} = 0.1 \text{ M}$. Since $\text{NaOH}$ is a strong base, it completely dissociates into $\text{Na}^+$ and $\text{OH}^-$ ions in aqueous solution . Therefore, the concentration of hydroxyl ions, $[\text{OH}^-] = 0.1 \text{ M} = 10^{-1} \text{ M}$. We know the ionic product of water, $K_w = [\text{H}^+][\text{OH}^-] = 10^{-14}$ at $298 \text{ K}$ . Thus, $[\text{H}^+] = \frac{K_w}{[\text{OH}^-]} = \frac{10^{-14}}{10^{-1}} = 10^{-13} \text{ M}$. Hence, the $\text{H}^+$ ion concentration is $10^{-13} \text{ M}$.