Question

1. a 500.0 ml acid solution is 0.100 mol l⁻¹ in hcl and 0.400 mol l⁻¹ hno₃. what volume of a 0.150 mol l⁻¹ koh solution would completely neutralize this acid solution? a) 1.33 l b) 1.00 l c) 1.67 l d) 0.333 l e) 0.150 l

Explanation:

Step1: Calculate moles of H⁺ from HCl

The volume of the acid - solution is $V = 500.0\ mL=0.500\ L$. The concentration of HCl is $C_{HCl}=0.100\ mol/L$. Since HCl is a strong acid and dissociates completely as $HCl
ightarrow H^{+}+Cl^{-}$, the number of moles of $H^{+}$ from HCl, $n_{HCl}$ is given by the formula $n = C\times V$. So, $n_{HCl}=0.100\ mol/L\times0.500\ L = 0.0500\ mol$.

Step2: Calculate moles of H⁺ from HNO₃

The concentration of $HNO_{3}$ is $C_{HNO_{3}} = 0.400\ mol/L$. Since $HNO_{3}$ is a strong acid and dissociates completely as $HNO_{3}
ightarrow H^{+}+NO_{3}^{-}$, the number of moles of $H^{+}$ from $HNO_{3}$, $n_{HNO_{3}}$ is $n_{HNO_{3}}=0.400\ mol/L\times0.500\ L=0.200\ mol$.

Step3: Calculate total moles of H⁺

The total number of moles of $H^{+}$, $n_{total\ H^{+}}=n_{HCl}+n_{HNO_{3}}$. So, $n_{total\ H^{+}}=0.0500\ mol + 0.200\ mol=0.250\ mol$.

Step4: Determine moles of OH⁻ needed for neutralization

The reaction between $H^{+}$ and $OH^{-}$ is $H^{+}+OH^{-}
ightarrow H_{2}O$. The mole - ratio of $H^{+}$ to $OH^{-}$ is 1:1. So, the number of moles of $OH^{-}$ needed to neutralize the $H^{+}$ is $n_{OH^{-}} = n_{total\ H^{+}}=0.250\ mol$.

Step5: Calculate volume of KOH solution

The concentration of KOH is $C_{KOH}=0.150\ mol/L$. Using the formula $V=\frac{n}{C}$, where $n$ is the number of moles of KOH (equal to the number of moles of $OH^{-}$) and $C$ is the concentration of KOH. So, $V_{KOH}=\frac{0.250\ mol}{0.150\ mol/L}\approx1.67\ L$.

Answer:

C. 1.67 L