Question

partners name_ivy tran instructor_dr. jia goal molarity of the naoh solution (original concentration) 0.05 m (2 pts) molarity of hcl solutions: dilute 0.008226(1 pt) concentrated 0.03403 (1 pt) results and data: part 1 - original naoh solution grams of naoh needed to make 250 ml of the original naoh solution 0.34 (show calculation and give correct significant figures) (4 pts) molarity of hcl solution needed to titrate the original naoh solution (show calculations for the volume of hcl needed to titrate the original naoh solution using each of the given molarities of hcl solutions) (10 pts)

Explanation:

Step1: Calculate moles of NaOH

The molarity formula is $M=\frac{n}{V}$, where $M$ is molarity, $n$ is moles and $V$ is volume in liters. Given $M = 0.05\ M$ and $V=250\ mL = 0.25\ L$. Rearranging for $n$, we get $n = M\times V$. So $n=0.05\ mol/L\times0.25\ L = 0.0125\ mol$.

Step2: Calculate mass of NaOH

The molar mass of $NaOH$ is $M_{NaOH}=22.99\ g/mol + 16.00\ g/mol+ 1.01\ g/mol=40.00\ g/mol$. Using the formula $m = n\times M$, where $m$ is mass, $n$ is moles and $M$ is molar mass. So $m = 0.0125\ mol\times40.00\ g/mol = 0.50\ g$.

For the titration part, the reaction between $HCl$ and $NaOH$ is $HCl + NaOH=NaCl + H_2O$, so the mole - ratio of $HCl$ to $NaOH$ is $1:1$. Let the volume of $HCl$ be $V_{HCl}$ and its molarity be $M_{HCl}$, and the moles of $NaOH$ be $n_{NaOH}$. At the equivalence point $n_{HCl}=n_{NaOH}$.

For dilute HCl ($M_{HCl}=0.008226\ M$)

We know $n_{NaOH}=0.0125\ mol$. Since $n_{HCl}=n_{NaOH}$ and $n_{HCl}=M_{HCl}\times V_{HCl}$, then $V_{HCl}=\frac{n_{HCl}}{M_{HCl}}=\frac{0.0125\ mol}{0.008226\ mol/L}\approx1.52\ L$.

For concentrated HCl ($M_{HCl}=0.03403\ M$)

Again, since $n_{HCl}=n_{NaOH} = 0.0125\ mol$ and $n_{HCl}=M_{HCl}\times V_{HCl}$, then $V_{HCl}=\frac{n_{HCl}}{M_{HCl}}=\frac{0.0125\ mol}{0.03403\ mol/L}\approx0.367\ L$.

Answer:

The mass of $NaOH$ needed is $0.50\ g$.
For the dilute $HCl$ ($0.008226\ M$), the volume of $HCl$ needed is approximately $1.52\ L$.
For the concentrated $HCl$ ($0.03403\ M$), the volume of $HCl$ needed is approximately $0.367\ L$.