An analyst wants to convert $1 \mathrm{~L} \mathrm{~HCl}$ of $\mathrm{pH}=1$ to a solution of $\mathrm{HCl}$ of $\mathrm{pH} ~2$. The volume of water needed to do this dilution is __________ $\mathrm{mL}$. (Nearest integer)

<p>We can use the formula for pH to calculate the concentration of hydrogen ions in each solution:</p> <p>$\mathrm{pH} = -\log_{10}[\mathrm{H}^+]$</p> <p>For the first solution, we have:</p> <p>$1 = -\log_{10}[\mathrm{H}^+]$</p> <p>Solving for $[\mathrm{H}^+]$, we get:</p> <p>$[\mathrm{H}^+] = 0.1 \mathrm{~M}$</p> <p>For the second solution, we want:</p> <p>$2 = -\log_{10}[\mathrm{H}^+]$</p> <p>Solving for $[\mathrm{H}^+]$, we get:</p> <p>$[\mathrm{H}^+] = 0.01 \mathrm{~M}$</p> <p>To dilute the first solution to the desired concentration, we can use the dilution equation:</p> <p>$C_1V_1 = C_2V_2$</p> <p>where $C_1$ and $V_1$ are the initial concentration and volume, and $C_2$ and $V_2$ are the final concentration and volume.</p> <p>We know $C_1 = 0.1 \mathrm{~M}$, $C_2 = 0.01 \mathrm{~M}$, and $V_1 = 1 \mathrm{~L}$. Solving for $V_2$, we get:</p> <p>$$V_2 = \frac{C_1V_1}{C_2} = \frac{(0.1 \mathrm{~M})(1 \mathrm{~L})}{0.01 \mathrm{~M}} = 10 \mathrm{~L}$$</p> <p>Therefore, we need to add $10-1=9$ liters of water to the initial solution to obtain the desired pH. Converting liters to milliliters, we get:</p> <p>$9 \mathrm{~L} \times \frac{1000 \mathrm{~mL}}{1 \mathrm{~L}} = 9000 \mathrm{~mL}$</p> <p>So the volume of water needed is 9000 mL or 9,000 mL (nearest integer).</p>