Suppose 25.00 mL of 0.0500 M CH3NH2 (Kb = 4.4 × 10-4) is titrated with 0.0625...

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Chemistry

Suppose 25.00 mL of 0.0500 M CH3NH2(Kb = 4.4 × 10-4) is titrated with 0.0625 MHCl.

How many millimoles of CH3NH2 are presentinitially?

How many millimoles of HCl are required to reach the equivalencepoint?

What is the equivalence point (in mL)?

What is the total volume (in mL) of analyte solution at theequivalence point?

At the equivalence point, which are the principal speciespresent (excluding Cl-)?

Select one:

a. H2O and CH3NH2

b. H2O andCH3NH3+

c. H2O, H3O+ andCH3NH3+

d. H2O, H3O+ andCH3NH2

e. H2O, H3O+,CH3NH2, andCH3NH3+

What is the pH at the equivalence point?

Answer & Explanation Solved by verified expert
4.2 Ratings (750 Votes)
We are using 25 mL of 005 M CH3NH2 millimols of CH3NH2 25 mL 1 L1000 mL 005 molL 1000 millimol 1 mol 125 millimol millimols of CH3NH2 present initially are 125 The chemical equation for the reaction between CH3NH2 and HCl is given as CH3NH2 HCl CH3NH3Cl From the above equation we can say that the stoichiometry for CH3NH2 and HCl is 11 Therefore millimols of HCl required to reach    See Answer
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