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Consider the titration of a 24.0 ml sample of 0.115 moll-1 ch3nh2 (kb=4.4×10-4) with 0.155 moll-1 hbr. Determine each quantity:

User Leifparker
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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.

User DLiKS
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