Chemistry Practice Problems Weak Acid Strong Base Titrations Practice Problems Solution: 1.00 g CH3NH3Cl (MW = 67.52 g/mol) and 1.00 g Ba(O...

Solution: 1.00 g CH3NH3Cl (MW = 67.52 g/mol) and 1.00 g Ba(OH) 2∙8H2O (a strong base, MW = 315.47 g/mol) are added to 80 mL of water and dissolved. The resulting solution is diluted to 100 mL with water at 25°C.Write the governing equilibrium.A.  CH3NH2(aq) + H2O(l) ⇌ CH3NH3+(aq) + OH-(aq)B. CH3NH3Cl(aq) ⇌ CH3NH3+(aq) + Cl-(aq)C. Ba(OH)2(aq) ⇌ Ba2+(aq) + OH-(aq) D. 2 CH3NH3Cl(aq) + Ba(OH)2(aq) ⇌ 2 CH3NH2(aq) + BaCl2 + 2 H2O(l)E. CH3NH3+(aq) + OH-(aq)  ⇌ CH3NH2(aq) + H2O(l)

Problem

1.00 g CH3NH3Cl (MW = 67.52 g/mol) and 1.00 g Ba(OH) 2∙8H2O (a strong base, MW = 315.47 g/mol) are added to 80 mL of water and dissolved. The resulting solution is diluted to 100 mL with water at 25°C.

Write the governing equilibrium.

A.  CH3NH2(aq) + H2O(l) ⇌ CH3NH3+(aq) + OH-(aq)

B. CH3NH3Cl(aq) ⇌ CH3NH3+(aq) + Cl-(aq)

C. Ba(OH)2(aq) ⇌ Ba2+(aq) + OH-(aq) 

D. 2 CH3NH3Cl(aq) + Ba(OH)2(aq) ⇌ 2 CH3NH2(aq) + BaCl2 + 2 H2O(l)

E. CH3NH3+(aq) + OH-(aq)  ⇌ CH3NH2(aq) + H2O(l)

Solution

We're being asked to determine the governing equilibrium for the given solution.


CH3NH3Cl is a salt and will break up into its ions in the solution:

CH3NH3Cl(aq) → CH3NH3+(aq) + Cl-(aq) 

CH3NH3+(aq) → positively charged amine → weak acid
Cl-(aq) → ion from a strong acid (HCl) → neutral ion and will not contribute to pH

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