Final answer:
To titrate a 24.0 mL sample of 0.115 mol L-1 CH3NH2 with 0.155 mol L-1 HBr, 17.8 mL of HBr is required.
Step-by-step explanation:
The question involves calculating the volume of 0.155 mol L-1 HBr required to titrate a 24.0 mL sample of 0.115 mol L-1 CH3NH2 (methylamine). The titration can be thought of as an acid-base reaction where HBr, a strong acid, reacts with CH3NH2, a weak base.
The reaction can be represented as:
CH3NH2 + HBr → CH3NH3+ + Br-
To solve for the volume of HBr needed, we must first calculate the moles of CH3NH2 in the 24.0 mL sample.
moles of CH3NH2 = 24.0 mL × (1 L/1000 mL) × 0.115 mol L-1 = 0.00276 moles
Since HBr is a strong acid and reacts in a 1:1 molar ratio with CH3NH2, the same number of moles of HBr will be required for the neutralization:
moles of HBr = 0.00276 moles
Finally, we will calculate the volume of HBr solution needed using its concentration:
Volume of HBr = moles of HBr / concentration of HBr = 0.00276 moles / 0.155 mol L-1 = 0.0178 L or 17.8 mL
Therefore, 17.8 mL of 0.155 mol L-1 HBr is required to titrate the given sample of CH3NH2.