An analyst wants to convert 1LHCl of pH = 1 to a solution of HCl of...

The concentration of ${H}^{+}$ or say $HCl$ in both the given solution is ${10}^{– 1}$ and ${10}^{– 2}$ mole ${L}^{– 1}$ respectively.

For the given amount of $HCl$ the concentration depends over dilution hence

${M}_{1}{V}_{1} = {M}_{2}{V}_{2}\dfrac{{M}_{1}}{{M}_{2}} = \dfrac{{V}_{2}}{{V}_{1}}$

$\dfrac{{10}^{- 1}}{{10}^{- 2}} = \dfrac{{V}_{2}}{1L}$

${V}_{2} = 10L = 10000mL$

$∆ V = 10000 - 1000mL = 9000mL$