Question

calculate the volume of 1.50 × 10⁻² m naoh that must be added to 500.0 ml of 0.200 m hcl to give a solution that has ph = 2.05. volume = l submit answer try another version 3 item attempts remaining

Explanation:

Step1: Find $[H^+]$ from pH

Given $pH = 2.05$, use $[H^+]=10^{-pH}$. So $[H^+]=10^{-2.05}\approx8.91\times10^{-3}\ M$.

Step2: Define moles of HCl and NaOH

Let $V$ (in L) be volume of NaOH. Moles of HCl: $n_{HCl}=0.200\ M\times0.500\ L = 0.100\ mol$. Moles of NaOH: $n_{NaOH}=1.50\times10^{-2}\ M\times V = 0.015V\ mol$.

Step3: Moles of $H^+$ after reaction

After reaction, $[H^+]=\frac{n_{HCl}-n_{NaOH}}{V_{total}}$, where $V_{total}=0.500 + V$ (in L). So $8.91\times10^{-3}=\frac{0.100 - 0.015V}{0.500 + V}$.

Step4: Solve for V

Multiply both sides by $(0.500 + V)$: $8.91\times10^{-3}(0.500 + V)=0.100 - 0.015V$. Expand: $0.004455 + 0.00891V = 0.100 - 0.015V$. Combine like terms: $0.00891V + 0.015V = 0.100 - 0.004455$. $0.02391V = 0.095545$. Then $V=\frac{0.095545}{0.02391}\approx3.996\ L\approx4.00\ L$.

Answer:

$\approx 4.00$