Question

report sheet
name asha
partners name lily tran
date
section 43
instructor dr. tiu
how much solute is in your solution?
goal molarity of the naoh solution (original concentration) 0.03m
molarity of hcl solutions
dilute 0.00822m
concentrated 0.03403m
results and data:
part 1 - original naoh solution
grams of naoh needed to make 250 ml of the original naoh solution (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)
original naoh solution\ttitration 1\ttitration 2
initial buret reading (hcl)\t25.16 ml\t34.82 ml
final buret reading (hcl)\t34.82 ml\t43.92 ml
total volume of hcl added\t9.66 ml\t9.10 ml
concentration of the naoh solution based on the titration data\t0.013127 m\t0.012175 m
average concentration of the concentrated naoh solution 0.03m

Explanation:

Step1: Calculate moles of NaOH for 250 mL of 0.03 M solution

Use the formula $n = M\times V$, where $M = 0.03\ M$ and $V=0.250\ L$ (since $250\ mL = 0.250\ L$). So $n = 0.03\ mol/L\times0.250\ L= 0.0075\ mol$.

Step2: Calculate mass of NaOH

The molar - mass of $NaOH$ is $M_{NaOH}=22.99 + 16.00+1.01=40.00\ g/mol$. Using the formula $m = n\times M$, we have $m = 0.0075\ mol\times40.00\ g/mol = 0.3\ g$.

Step3: Calculate volume of HCl for titration

The reaction between $HCl$ and $NaOH$ is $HCl + NaOH=NaCl + H_2O$, so the mole - ratio of $HCl$ to $NaOH$ is $1:1$. Let $V_{NaOH}$ be the volume of $NaOH$ and $M_{NaOH}$ be its molarity, and $V_{HCl}$ be the volume of $HCl$ and $M_{HCl}$ be its molarity. Then $n_{NaOH}=n_{HCl}$, or $M_{NaOH}V_{NaOH}=M_{HCl}V_{HCl}$. Given $V_{NaOH}$ (assume $V_{NaOH} = 10\ mL = 0.010\ L$), $M_{NaOH}=0.03\ M$, and $M_{HCl}=0.03403\ M$, we can solve for $V_{HCl}$: $V_{HCl}=\frac{M_{NaOH}V_{NaOH}}{M_{HCl}}=\frac{0.03\ mol/L\times0.010\ L}{0.03403\ mol/L}\approx8.8\ mL$.

Step4: Calculate concentration of NaOH from titration data

For titration 1:
Using $M_{NaOH}V_{NaOH}=M_{HCl}V_{HCl}$, we know $V_{HCl} = 9.64\ mL = 0.00964\ L$, assume $V_{NaOH}=10\ mL = 0.010\ L$ and $M_{HCl}=0.03403\ M$. Then $M_{NaOH}=\frac{M_{HCl}V_{HCl}}{V_{NaOH}}=\frac{0.03403\ mol/L\times0.00964\ L}{0.010\ L}=0.0328\ M$ (this seems inconsistent with the given value in the table, but using the formula).
For titration 2:
$V_{HCl}=9.10\ mL = 0.00910\ L$, assume $V_{NaOH}=10\ mL = 0.010\ L$ and $M_{HCl}=0.03403\ M$. Then $M_{NaOH}=\frac{M_{HCl}V_{HCl}}{V_{NaOH}}=\frac{0.03403\ mol/L\times0.00910\ L}{0.010\ L}=0.0310\ M$ (also inconsistent with the given value in the table).

Step5: Calculate average concentration

The average concentration of $NaOH$ based on the two titrations (using the values in the table) is $\frac{0.013127\ M + 0.012875\ M}{2}=0.013001\ M$ (but the given average is $0.03\ M$).

Answer:

• Mass of $NaOH$ for 250 mL of 0.03 M solution: $0.3\ g$
• Volume of $HCl$ for titration (using $M_{HCl}=0.03403\ M$ and $V_{NaOH} = 10\ mL$): $8.8\ mL$
• Concentration of $NaOH$ from titration 1 (using formula): $0.0328\ M$ (inconsistent with table value)
• Concentration of $NaOH$ from titration 2 (using formula): $0.0310\ M$ (inconsistent with table value)
• Average concentration of $NaOH$ (using table values): $0.013001\ M$ (different from given average of $0.03\ M$)