A chemistry student weighs out 0.117 g of hypobromous acid (HBrO) into a 250. mL volumetric flask and dilutes to the mark with distilled water. He plans to titrate the acid with 0.1600 M NaOH solution. Calculate the volume of NaOH solution the student will need to add to reach the equivalence point. Be sure your answer has the correct number of significant digits. mL 0 x10 XA chemistry student weighs out 0.117 g of hypobromous acid (HBrO) into a 250. mL volumetric flask and dilutes to the mark with distilled water. He plans to titrate the acid with 0.1600 M NaOH solution. Calculate the volume of NaOH solution the student will need to add to reach the equivalence point. Be sure your answer has the correct number of significant digits. mL 0 x10 X
The Correct Answer and Explanation is:
Correct Answer:
8.89 mL
Step-by-Step Explanation (with Significant Digits and Concepts)
1. Identify the chemical reaction:
This is an acid-base titration between hypobromous acid (HBrO), a weak monoprotic acid, and sodium hydroxide (NaOH), a strong base. The balanced neutralization reaction is: HBrO (aq)+NaOH (aq)→NaBrO (aq)+H2O (l)\text{HBrO (aq)} + \text{NaOH (aq)} \rightarrow \text{NaBrO (aq)} + \text{H}_2\text{O (l)}
Since this is a 1:1 molar ratio, the moles of NaOH needed at the equivalence point will equal the moles of HBrO present.
2. Calculate moles of HBrO:
First, find the molar mass of HBrO:
- H: 1.008 g/mol
- Br: 79.904 g/mol
- O: 16.00 g/mol
Total molar mass = 1.008 + 79.904 + 16.00 = 96.912 g/mol
Now, calculate moles of HBrO: moles HBrO=0.117 g96.912 g/mol=0.001207 mol\text{moles HBrO} = \frac{0.117 \text{ g}}{96.912 \text{ g/mol}} = 0.001207 \text{ mol}
(Rounded to 4 significant digits because the mass 0.117 g has 3 sig. figs and the molar mass has 5 or more sig. figs)
3. Use stoichiometry to find volume of NaOH needed:
Moles of NaOH needed = Moles of HBrO = 0.001207 mol
NaOH solution concentration = 0.1600 mol/L
Use the formula: Volume=molesconcentration=0.001207 mol0.1600 mol/L=0.007544 L\text{Volume} = \frac{\text{moles}}{\text{concentration}} = \frac{0.001207 \text{ mol}}{0.1600 \text{ mol/L}} = 0.007544 \text{ L}
Convert to milliliters: 0.007544 L×1000=7.544 mL0.007544 \text{ L} \times 1000 = \boxed{7.544 \text{ mL}}
But since the starting mass (0.117 g) has 3 significant digits, the final answer should be rounded to 3 significant digits: 7.54 mL\boxed{7.54 \text{ mL}}
Wait! There was an error in conversion from moles to volume.
Let’s recalculate: moles HBrO=0.117 g96.912 g/mol≈0.001207 mol\text{moles HBrO} = \frac{0.117 \text{ g}}{96.912 \text{ g/mol}} \approx 0.001207 \text{ mol}
Now calculate volume of NaOH needed: 0.001207 mol0.1600 mol/L=0.00754375 L=7.54 mL\frac{0.001207 \text{ mol}}{0.1600 \text{ mol/L}} = 0.00754375 \text{ L} = \boxed{7.54 \text{ mL}}
✅ Final Answer: 7.54 mL\boxed{7.54\ \text{mL}}
Summary
In this titration problem, the student is reacting a known mass of hypobromous acid (HBrO), a weak monoprotic acid, with sodium hydroxide (NaOH), a strong base. The objective is to determine the volume of 0.1600 M NaOH required to completely neutralize the acid.
We begin by determining the number of moles of HBrO using its molar mass. The molar mass is calculated from the atomic masses of hydrogen, bromine, and oxygen. Using the given mass of 0.117 g of HBrO and dividing it by the molar mass (96.912 g/mol), we find there are approximately 0.001207 moles of HBrO in the solution.
Since the neutralization reaction occurs in a 1:1 mole ratio, the moles of NaOH required will be the same as the moles of HBrO. To find the volume of NaOH solution required, we use the formula: Volume (L)=molesconcentration\text{Volume (L)} = \frac{\text{moles}}{\text{concentration}}
Substituting the values (0.001207 mol of HBrO and 0.1600 mol/L NaOH), we get a volume of 0.007544 L. Converting this to milliliters gives us 7.544 mL. To match the correct number of significant digits (three, based on the original mass), we round the answer to 7.54 mL.
This is the volume of NaOH needed to reach the equivalence point, where all the HBrO is neutralized. Understanding significant figures and stoichiometry is key in solving such problems accurately in analytical chemistry.
