11. calculate the volume in milliliters of a 2.35 m naoh solution required to titrate the following…

Explanation:

Step1: Write the balanced chemical equations

(a) $HCl + NaOH=NaCl + H_2O$, mole - ratio of $HCl$ to $NaOH$ is 1:1. (b) $H_2SO_4+2NaOH = Na_2SO_4 + 2H_2O$, mole - ratio of $H_2SO_4$ to $NaOH$ is 1:2.

Step2: Use the formula $n = M\times V$ (where $n$ is the number of moles, $M$ is molarity and $V$ is volume)

(a) For $HCl$, $n_{HCl}=M_{HCl}\times V_{HCl}$, with $M_{HCl} = 2.43\ M$ and $V_{HCl}=50.00\ mL = 0.05000\ L$. So $n_{HCl}=2.43\ mol/L\times0.05000\ L = 0.1215\ mol$. Since $n_{NaOH}=n_{HCl}$ (from the balanced equation), and $M_{NaOH}=2.35\ M$, using $V_{NaOH}=\frac{n_{NaOH}}{M_{NaOH}}$, we have $V_{NaOH}=\frac{0.1215\ mol}{2.35\ mol/L}=0.0517\ L = 51.7\ mL$. (b) For $H_2SO_4$, $n_{H_2SO_4}=M_{H_2SO_4}\times V_{H_2SO_4}$, with $M_{H_2SO_4}=4.50\ M$ and $V_{H_2SO_4}=50.00\ mL = 0.05000\ L$. So $n_{H_2SO_4}=4.50\ mol/L\times0.05000\ L = 0.225\ mol$. Since $n_{NaOH} = 2n_{H_2SO_4}$ (from the balanced equation), $n_{NaOH}=2\times0.225\ mol = 0.450\ mol$. Using $V_{NaOH}=\frac{n_{NaOH}}{M_{NaOH}}$, with $M_{NaOH}=2.35\ M$, we get $V_{NaOH}=\frac{0.450\ mol}{2.35\ mol/L}=0.1915\ L = 191.5\ mL$.

Answer:

(a) 51.7 mL (b) 191.5 mL