Calculating the change in pH upon diluting an acid
As Ivan said in the comments, $\ce{HCl}$ is a strong acid. It dissociates as follows,
$$\ce{HCl -&gt; H+ + Cl-}$$
Now, let us assume the intial pH to be x. Since, $$\mathrm{pH} = \ce{-log[H+]}$$ $\ce{[H+]_i}$ = $\mathrm{10^{-x}}$, where subscript i means initial. Likewise, since final pH= x+1, $\ce{[H+]_f}$ = $\mathrm{10^{-(x+1)}}$ where f means final. Since $\mathrm{V_i = 400 mL}$, since total moles remain the same,
$$\mathrm{M_iV_i = M_fV_f}$$
we get $\mathrm{V_f = 4000 mL}$.
Hence amount of water to be added is $\mathrm{4000-400 = 3600 mL}$.